Introduction The heights of these radish plants are continuous random variables. (credit: modification of work “Radish (Raphanus sativus): White rust caused by Albugo candida” by Scot Nelson/ Flickr, Public Domain) Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a long distance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, rates of return from an investment, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables, as do all areas of risk analysis. NOTE The values of discrete and continuous random variables can be ambiguous. For example, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete random variable. You count the miles. If X is the distance you drive to work, then you measure values of X and X is a continuous random variable. For a second example, if X is equal to the number of books in a backpack, then X is a discrete random variable. If X is the weight of a book, then X is a continuous random variable because weights are measured. How the random variable is defined is very important.

Math


Properties of Continuous Probability Density Functions The graph of a continuous probability distribution is a curve. Probability is represented by area under the curve. We have already met this concept when we developed relative frequencies with histograms in Descriptive Statistics . The relative area for a range of values was the probability of drawing at random an observation in that group. Again with the Poisson distribution in Random Discrete Variables , the graph in Example 4.15 used boxes to represent the probability of specific values of the random variable. In this case, we were being a bit casual because the random variables of a Poisson distribution are discrete, whole numbers, and a box has width. Notice that the horizontal axis, the random variable x, purposefully did not mark the points along the axis. The probability of a specific value of a continuous random variable will be zero because the area under a point is zero. Probability is area. The curve is called the probability density function (abbreviated as pdf ). We use the symbol f ( x ) to represent the curve. f ( x ) is the function that corresponds to the graph; we use the density function f ( x ) to draw the graph of the probability distribution. Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf ). The cumulative distribution function is used to evaluate probability as area. Mathematically, the cumulative probability density function is the integral of the pdf, and the probability between two values of a continuous random variable will be the integral of the pdf between these two values: the area under the curve between these values. Remember that the area under the pdf for all possible values of the random variable is one, certainty. Probability thus can be seen as the relative percent of certainty between the two values of interest. The outcomes are measured, not counted. The entire area under the curve and above the x-axis is equal to one. Probability is found for intervals of x values rather than for individual x values. P(c < x < d) is the probability that the random variable X is in the interval between the values c and d . P(c < x < d) is the area under the curve, above the x -axis, to the right of c and the left of d . P(x = c) = 0 The probability that x takes on any single individual value is zero. The area below the curve, above the x -axis, and between x = c and x = c has no width, and therefore no area (area = 0). Since the probability is equal to the area, the probability is also zero. P(c < x < d) is the same as P(c ≤ x ≤ d) because probability is equal to area. We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, integral calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, the formulas were found by using the techniques of integral calculus. There are many continuous probability distributions. When using a continuous probability distribution to model probability, the distribution used is selected to model and fit the particular situation in the best way. In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions. The graph shows a Uniform Distribution with the area between x = 3 and x = 6 shaded to represent the probability that the value of the random variable X is in the interval between three and six. The graph shows an Exponential Distribution with the area between x = 2 and x = 4 shaded to represent the probability that the value of the random variable X is in the interval between two and four. The graph shows the Standard Normal Distribution with the area between x = 1 and x = 2 shaded to represent the probability that the value of the random variable X is in the interval between one and two. For continuous probability distributions, PROBABILITY = AREA. Consider the function f ( x ) = 1 20 for 0 ≤ x ≤ 20. x = a real number. The graph of f ( x ) = 1 20 is a horizontal line. However, since 0 ≤ x ≤ 20, f ( x ) is restricted to the portion between x = 0 and x = 20, inclusive. f ( x ) = 1 20 for 0 ≤ x ≤ 20. The graph of f ( x ) = 1 20 is a horizontal line segment when 0 ≤ x ≤ 20. The area between f ( x ) = 1 20 where 0 ≤ x ≤ 20 and the x -axis is the area of a rectangle with base = 20 and height = 1 20 . AREA = 20 ( 1 20 ) = 1 Suppose we want to find the area between f( x ) = 1 20 and the x -axis where 0 < x < 2. AREA = ( 2 – 0 ) ( 1 20 ) = 0.1 ( 2 – 0 ) = 2 = base of a rectangle REMINDER area of a rectangle = (base)(height). The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written mathematically as P (0 < x < 2) = P ( x < 2) = 0.1. Suppose we want to find the area between f ( x ) = 1 20 and the x -axis where 4 < x < 15. AREA = ( 15 – 4 ) ( 1 20 ) = 0.55 ( 15 – 4 ) = 11 = the base of a rectangle The area corresponds to the probability P (4 < x < 15) = 0.55. Suppose we want to find P ( x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P ( x = 15) = (base)(height) = (0) ( 1 20 ) = 0 P ( X ≤ x ), which can also be written as P ( X < x ) for continuous distributions, is called the cumulative distribution function or CDF. Notice the "less than or equal to" symbol. We can also use the CDF to calculate P ( X > x ). The CDF gives "area to the left" and P ( X > x ) gives "area to the right." We calculate P ( X > x ) for continuous distributions as follows: P ( X > x ) = 1 – P ( X < x ). Label the graph with f ( x ) and x . Scale the x and y axes with the maximum x and y values. f ( x ) = 1 20 , 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. P ( 2.3 < x < 12.7 ) = ( base ) ( height ) = ( 12.7 − 2.3 ) ( 1 20 ) = 0.52 Try It Consider the function f ( x ) = 1 8 for 0 ≤ x ≤ 8. Draw the graph of f ( x ) and find P (2.5 < x < 7.5). P (2.5 < x < 7.5) = 0.625 Chapter Review The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P ( a < x < b ). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f ( x ), is used to draw the graph of the probability distribution. The total area under the graph of f ( x ) is one. The area under the graph of f ( x ) and between values a and b gives the probability P ( a < x < b ). The cumulative distribution function (cdf) of X is defined by P ( X ≤ x ). It is a function of x that gives the probability that the random variable is less than or equal to x . Formula Review Probability density function (pdf) f ( x ): f ( x ) ≥ 0 The total area under the curve f ( x ) is one. Cumulative distribution function (cdf): P ( X ≤ x ) Which type of distribution does the graph illustrate? Uniform Distribution Which type of distribution does the graph illustrate? Which type of distribution does the graph illustrate? Normal Distribution What does the shaded area represent? P (___< x < ___) What does the shaded area represent? P (___< x < ___) P (6 < x < 7) For a continuous probability distribution, 0 ≤ x ≤ 15. What is P ( x > 15)? What is the area under f ( x ) if the function is a continuous probability density function? one For a continuous probability distribution, 0 ≤ x ≤ 10. What is P ( x = 7)? A continuous probability function is restricted to the portion between x = 0 and 7. What is P ( x = 10)? zero f ( x ) for a continuous probability function is 1 5 , and the function is restricted to 0 ≤ x ≤ 5. What is P ( x < 0)? f ( x ), a continuous probability function, is equal to 1 12 , and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)? one Find the probability that x falls in the shaded area. Find the probability that x falls in the shaded area. 0.625 Find the probability that x falls in the shaded area. f ( x ), a continuous probability function, is equal to 1 3 and the function is restricted to 1 ≤ x ≤ 4. Describe P ( x > 3 2 ) . The probability is equal to the area from x = 3 2 to x = 4 above the x-axis and up to f ( x ) = 1 3 . Homework For each probability and percentile problem, draw the picture. Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes” or “no.” You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor. What part of the experiment will yield discrete data? What part of the experiment will yield continuous data? When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why? Age is a measurement, regardless of the accuracy used. Uniform Distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b ; it is often referred as the rectangular distribution because the graph of the pdf has the form of a rectangle. The mean is μ = a + b 2 and the standard deviation is σ = ( b − a ) 2 12 . The probability density function is f ( x ) = 1 b − a for a < x < b or a ≤ x ≤ b . The cumulative distribution is P ( X ≤ x ) = x − a b − a . Exponential Distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital. The mean is μ = 1 m and the standard deviation is σ = 1 m . The probability density function is f ( x ) = m e - m x or f ( x ) = 1 μ e - 1 μ x , x ≥ 0 and the cumulative distribution function is P ( X ≤ x ) = 1 - e − m x or P ( X ≤ x ) = 1 - e − 1 μ x .

Math


The Uniform Distribution The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints. The mathematical statement of the uniform distribution is f ( x ) = 1 b − a for a ≤ x ≤ b where a = the lowest value of x and b = the highest value of x . Formulas for the theoretical mean and standard deviation are μ = a + b 2 and σ = ( b − a ) 2 12 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. a. What is the probability that a person waits fewer than 12.5 minutes? a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U (0, 15). Write the probability density function. f ( x ) = 1 15 − 0 = 1 15 for 0 ≤ x ≤ 15. Find P ( x < 12.5). Draw a graph. P ( x < k ) = ( base ) ( height ) = ( 12.5 - 0 ) ( 1 15 ) = 0.8333 The probability a person waits less than 12.5 minutes is 0.8333. b. On the average, how long must a person wait? Find the mean, μ , and the standard deviation, σ . b. μ = a + b 2 = 15 + 0 2 = 7.5. On the average, a person must wait 7.5 minutes. σ = ( b - a ) 2 12 = ( 15 - 0 ) 2 12 = 4.3. The Standard deviation is 4.3 minutes. c. Ninety percent of the time, the time a person must wait falls below what value? NOTE This asks for the 90 th percentile. c. Find the 90 th percentile. Draw a graph. Let k = the 90 th percentile. P ( x < k ) = ( base ) ( height ) = ( k − 0 ) ( 1 15 ) 0.90 = ( k ) ( 1 15 ) k = ( 0.90 ) ( 15 ) = 13.5 The 90 th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. Try It The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive. Find a and b and describe what they represent. Write the distribution. Find the mean and the standard deviation. What is the probability that the duration of games for a team in a single season is between 480 and 500 hours? a is 447, and b is 521. a is the minimum duration of games for a team for the 2011 season, and b is the maximum duration of games for a team for the 2011 season. X ~ U (447, 521). μ = 484, and σ = 21.36 P (480 < x < 500) = 0.2703 Chapter Review If X has a uniform distribution where a < x < b or a ≤ x ≤ b , then X takes on values between a and b (may include a and b ). All values x are equally likely. We write X ∼ U ( a , b ). The mean of X is μ = a + b 2 . The standard deviation of X is σ = ( b − a ) 2 12 . The probability density function of X is f ( x ) = 1 b − a for a ≤ x ≤ b . The cumulative distribution function of X is P ( X ≤ x ) = x − a b − a . X is continuous. The probability P ( c < X < d ) may be found by computing the area under f ( x ), between c and d . Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. Formula Review X = a real number between a and b (in some instances, X can take on the values a and b ). a = smallest X ; b = largest X X ~ U (a, b) The mean is μ = a + b 2 The standard deviation is σ = ( b – a ) 2 12 Probability density function: f ( x ) = 1 b − a for a ≤ X ≤ b Area to the Left of x : P ( X < x ) = ( x – a ) ( 1 b − a ) Area to the Right of x : P ( X > x ) = ( b – x ) ( 1 b − a ) Area Between c and d : P ( c < x < d ) = (base)(height) = ( d – c ) ( 1 b − a ) pdf: f ( x ) = 1 b − a for a ≤ x ≤ b cdf: P ( X ≤ x ) = x − a b − a mean µ = a + b 2 standard deviation σ = ( b − a ) 2 12 P ( c < X < d ) = ( d – c ) ( 1 b – a ) References McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995. Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. 1.5 2.4 3.6 2.6 1.6 2.4 2.0 3.5 2.5 1.8 2.4 2.5 3.5 4.0 2.6 1.6 2.2 1.8 3.8 2.5 1.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6 The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as X ~ U (1.5, 4.5). What type of distribution is this? In this distribution, outcomes are equally likely. What does this mean? It means that the value of x is just as likely to be any number between 1.5 and 4.5. What is the height of f ( x ) for the continuous probability distribution? What are the constraints for the values of x ? 1.5 ≤ x ≤ 4.5 Graph P (2 < x < 3). What is P (2 < x < 3)? 0.3333 What is P (x < 3.5 | x < 4)? What is P ( x = 1.5)? zero Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. 0.6 Use the following information to answer the next eight exercises. A distribution is given as X ~ U (0, 12). What is a ? What does it represent? What is b ? What does it represent? b is 12, and it represents the highest value of x . What is the probability density function? What is the theoretical mean? six What is the theoretical standard deviation? Draw the graph of the distribution for P ( x > 9). Find P ( x > 9). Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. What is being measured here? In words, define the random variable X . X = The age (in years) of cars in the staff parking lot Are the data discrete or continuous? The interval of values for x is ______. 0.5 to 9.5 The distribution for X is ______. Write the probability density function. f ( x ) = 1 9 where x is between 0.5 and 9.5, inclusive. Graph the probability distribution. Sketch the graph of the probability distribution. Identify the following values: Lowest value for x – : _______ Highest value for x – : _______ Height of the rectangle: _______ Label for x -axis (words): _______ Label for y -axis (words): _______ Find the average age of the cars in the lot. μ = 5 Find the probability that a randomly chosen car in the lot was less than four years old. Sketch the graph, and shade the area of interest. Find the probability. P ( x < 4) = _______ Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. Sketch the graph, shade the area of interest. Find the probability. P ( x < 4 | x < 7.5) = _______ Answers may vary. 3.5 7 What has changed in the previous two problems that made the solutions different? Find the third quartile of ages of cars in the lot. This means you will have to find the value such that 3 4 , or 75%, of the cars are at most (less than or equal to) that age. Sketch the graph, and shade the area of interest. Find the value k such that P ( x < k ) = 0.75. The third quartile is _______ Answers may vary. k = 7.25 7.25 Homework For each probability and percentile problem, draw the picture. Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks). Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that a person is born at the exact moment week 19 starts. That is, find P ( x = 19) = _________ P (2 < x < 31) = _________ Find the probability that a person is born after week 40. P (12 < x | x < 28) = _________ A random number generator picks a number from one to nine in a uniform manner. Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ P (3.5 < x < 7.25) = _________ P ( x > 5.67) P ( x > 5 | x > 3) = _________ Answers may vary. f ( x ) = 1 8 where 1 ≤ x ≤ 9 five 2.3 15 32 333 800 2 3 According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. Define the random variable. X = _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that the individual lost more than ten pounds in a month. Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 12 pounds in the month. P (7 < x < 13 | x > 9) = __________. State this in a probability question, similarly to parts g and h, draw the picture, and find the probability. A subway train on the Red Line arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. Define the random variable. X = _______ Graph the probability distribution. f ( x ) = _______ μ = _______ σ = _______ Find the probability that the commuter waits less than one minute. Find the probability that the commuter waits between three and four minutes. X represents the length of time a commuter must wait for a train to arrive on the Red Line. Graph the probability distribution. f ( x ) = 1 8 where 0 ≤ x ≤ 8 four 2.31 1 8 1 8 The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class. Define the random variable. X = _________ Graph the probability distribution. f ( x ) = _________ μ = _________ σ = _________ Find the probability that she is over 6.5 years old. Find the probability that she is between four and six years old. Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution. What is the average waiting time (in minutes)? zero two three four d The probability of waiting more than seven minutes given a person has waited more than four minutes is? 0.125 0.25 0.5 0.75 b The time (in minutes) until the next bus departs a major bus depot follows a distribution with f ( x ) = 1 20 where x goes from 25 to 45 minutes. Define the random variable. X = ________ Graph the probability distribution. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous). μ = ________ σ = ________ Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. P (25 < x < 55) = _________. State this in a probability statement, similarly to parts g and h, draw the picture, and find the probability. Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution. Find the probability that the value of the stock is more than $19. Find the probability that the value of the stock is between $19 and $22. Given that the stock is greater than $18, find the probability that the stock is more than $21. The probability density function of X is 1 25 − 16 = 1 9 . P ( X > 19) = (25 – 19) ( 1 9 ) = 6 9 = 2 3 . P (19 < X < 22) = (22 – 19) ( 1 9 ) = 3 9 = 1 3 . This is a conditional probability question. P(x > 21 | x > 18). You can do this two ways: Draw the graph where a is now 18 and b is still 25. The height is 1 ( 25 − 18 ) = 1 7 So, P ( x > 21 | x > 18) = (25 – 21) ( 1 7 ) = 4/7. Use the formula: P ( x > 21 | x > 18) = P ( x > 21 ∩ x > 18 ) P ( x > 18 ) = P ( x > 21 ) P ( x > 18 ) = ( 25 − 21 ) ( 25 − 18 ) = 4 7 . A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution. Find the average time between fireworks. Find probability that the time between fireworks is greater than four seconds. The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. Find the probability that the truck driver goes more than 650 miles in a day. Find the probability that the truck drivers goes between 400 and 650 miles in a day. P ( X > 650) = 700 − 650 700 − 300 = 50 400 = 1 8 = 0.125 P (400 < X < 650) = 650 − 400 700 − 300 = 250 400 = 0.625 Conditional Probability the likelihood that an event will occur given that another event has already occurred.

Math


The Exponential Distribution The exponential probability density function is built upon the general exponential function where the variable is an exponent: f ( x ) = a ( b ) x . This equation can be converted to a natural system of logarithms with a base e that has an approximate value of 2.71828. The exponential probability density function is valuable with a number of practical applications. The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length of time, in minutes, of long-distance business telephone calls and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution. Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, marketing studies have shown that the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money. Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts. A firm will use the exponential distribution analysis to set the number of months they will provide a warranty based upon the probability of a specific time until the failure of the product. The random variable for the exponential distribution is continuous and often measures a passage of time, although it can be used in other continuous random variable applications. Typical questions may be “What is the probability that some event will occur within the next x hours or days?” or “What is the probability that some event will occur between x 1 hours and x 2 hours?” or “What is the probability that the event will take more than x 1 hours to perform?” In short, the random variable X equals ( a ) the time between events or ( b ) the passage of time to complete an action (e.g., wait on a customer). Be sure to differentiate the exponential distribution random variable and that of the Poisson distribution. The Poisson has a discrete random variable that gives us the probability that x number of occurrences will occur in the next time period. The continuous exponential distribution provides us the probability that an X -long time period will occur between now and the next occurrence. We will explore more this link between the exponential distribution and the Poisson distribution. The exponent probability density function is given by: f ( x ) = 1 μ e - x μ where μ is the historical average waiting time. The exponential distribution has a mean and standard deviation both equal to μ . To find a probability with a continuous random variable, the cumulative distribution is integrated between the relevant values of x . The definite integral of a function provides the area under the function between two values of x . For the exponential distribution, there are three cases that depend upon the question asked. Case I would answer the question “What is the probability that the time to completion is les s than X ?” The probability question asked would be written as P ( X ≤ x ) . For example, to calculate the probability that a customer will be served within the next hour given that the historical waiting time was 30 minutes, we integrate the exponential probability distribution function 1 μ e - x μ from x = 0 to x = 60 . F ( x ) = P ( X ≤ x ) = ∫ 0 x 1 μ e - x μ d x = 1 - e - x μ Case I graphs the exponential probability density function for this example. The area under the function from zero to 60 minutes is the probability for the question asked. Note that the exponential probability distribution at zero is at the value of 1 / µ . Solving for the probability density function equation above where µ = 30 is the historical waiting time, and x = 60 , which is the time we are interested in. P ( x ≤ 60 ) = 1 - e - x μ = 1 - e - 60 30 = 0 . 865 With a historical waiting time of 30 minutes, certainly we would expect the high probability of 0.865 of being waited on in an hour’s time. (A probability of 0.865 is almost a certainty.) Case II would answer questions as “What is the probability more than X ?” What is the probability that a person will have to wait MORE THAN x minutes? The relevant probability density function would be P ( X ≥ x α ) = 1 - 1 - e - x μ = e - x μ and is presented in . Case II Imagine an electric component is guaranteed to last two years. If it fails within the two-year guarantee, the purchaser will receive their money back in full. The manufacture knows by careful monitoring of their products the historical life of this component is 750 days, or two-and-one-half years. Are they engaged in a risky business policy? Here we can calculate the probability that a component will last longer than 730 days (two years) and thus no payment is required by the firm. The historical mean is 750 days, and the relevant time period is x = 730 days. Using the exponential probability density function, it is found that the probability this electric component will last more than two years is only 0.378, as shown on . P ( X ≥ x ) = e - x μ = e - 730 750 = 0 . 378 Because the working life of each of the electric components is an independent random variable, we can say that approximately 38 percent of these components survive two years and thus 62 percent of purchasers of these components will be asking for a full refund. In short, the low production quality and thus high refund requirements will result in significant financial costs. We could expect that the two-year guarantee will be eliminated soon. Case III allows for knowing the probability that an occurrence will occur within a window of time. For example, when asking for the arrival time of an airplane, one might be given a range of time such as “plus or minus from the historical expected arrival time measured in hours from the time of take-off.” In this case, the mean is the historical expected arrival time given the historical travel time. Assume that the historical travel time between these two points is four hours. Case III requires calculating the probability between two points. : Case III shows the two points that are one hour longer and one hour sooner than the historical arrival time, which is midway between these points. If the historical travel time is four hours, then x o is three hours of travel time and x 1 is five hours of travel time. The mathematical expression is: P ( x 0 ≤ x ≤ x 1 ) = P ( X ≥ x 0 ) - P ( X ≥ x 1 ) = e - x 0 μ - e - x 1 μ P ( 3 ≤ x ≤ 5 ) = e - 3 4 - e - 5 4 = 0 . 186 The probability the flight will arrive plus or minus the historical arrival time of four hours from takeoff is 0.186. Intuitively, this seems a very small probability to hit such a broad window. At the core, the standard deviation of average flight time must be very large. Exponential Distribution and Decay Factor The core of the exponential distribution is e, the natural logarithm in an exponential function with the variable - x μ in the exponent. The exponent is negative, and thus it describes an exponentially declining function as we have seen in , , and above. This gives rise to an alternative formula for the exponential probability distribution: f ( x ) = m e - m F ( x ) = ∫ 0 x m e - m t d t = 1 - e - m x Where m = 1 μ is given the name decay factor and measures the speed of decay. The decay factor simply measures how rapidly the probability of an event declines as the random variable X increases. shows the exponential distribution for three different decay factors. As before, the intercept on the vertical axis, the probability density, is the decay factor. A decay factor of 1.5, as in the first exponential function, is three time the speed of the third exponential distribution with a decay factor of 0.5. Looking at the probability for each of the three functions at the same value of x 1 , it is clear that the first function has a significantly greater probability of occurrence at x 1 than the other two (i.e., P ( X α ≤ x 1 ) > P ( X b ≤ x 1 ) > P ( X c ≤ x 1 ) comparing the relevant probabilities for each of the functions a, b, c in order). Finally, the exponential distribution shows with drama the true meaning of “exponential.” In each unit, a change in x in an exponential distribution has a significant change in probability. For example, if the simple exponential distribution function Y = 2 x changes from x = 4 to x = 5 , the y -value increases from 16 to 32, but a one-unit change from x = 10 to x = 11 causes the y -value to increase from 1,024 to 2,048. Likewise, in an exponential function with a negative exponent such as the exponential probability distribution, the changes are startling for each unit change at higher levels of X . At X = – 4 to x = - 5 , the y -value changes from 0.063 to 0.031, while a one-unit change from X = – 10 to X = – 11 results in a change in the y -value from 0.00097 to 0.00048. The conclusion is that probabilities in an exponential probability distribution at large x -values are scarce almost to the point of nonexistence. Let X = amount of time (in minutes) a postal clerk spends with a customer. The time is known from historical data to have an average amount of time equal to four minutes. It is given that μ = 4 minutes, that is, the average time the clerk spends with a customer is 4 minutes. Remember that we are still doing probability and thus we have to be told the population parameters such as the mean. To do any calculations, we need to know the mean of the distribution: the historical time to provide a service, for example. Knowing the historical mean allows the calculation of the decay parameter, m . m = 1 μ . Therefore, m = 1 4 = 0.25 When the notation used the decay parameter, m , the probability density function is presented as f ( x ) = m e − m x , which is simply the original formula with m substituted for 1 μ , or f ( x ) = 1 μ e − 1 μ x . To calculate probabilities for an exponential probability density function, we need to use the cumulative density function. As shown below, the curve for the cumulative density function is: f ( x ) = 0.25 e –0.25 x where x is at least zero and m = 0.25. For example, f (5) = 0.25 e (-0.25)(5) = 0.072. In other words, the function has a value of .072 when x = 5. The graph is as follows: Notice the graph is a declining curve. When x = 0, f ( x ) = 0.25 e (−0.25)(0) = (0.25)(1) = 0.25 = m . The maximum value on the y -axis is always m , one divided by the mean. Try It The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution. X ~ Exp (0.125); f ( x ) = 0.125e –0.125 x ; a. Using the information in , find the probability that a clerk spends four to five minutes with a randomly selected customer. a. Find P (4 < x < 5). The cumulative distribution function (CDF) gives the area to the left. P ( x < x ) = 1 – e –mx P ( x < 5) = 1 – e (–0.25)(5) = 0.7135 and P ( x < 4) = 1 – e (–0.25)(4) = 0.6321 P (4 < x < 5)= 0.7135 – 0.6321 = 0.0814 Try It The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait? P ( x < 10) = 0.4866 50 th percentile = 10.40 On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed. a. What is the probability that a computer part lasts more than 7 years? a. Let x = the amount of time (in years) a computer part lasts. μ = 10 so m = 1 μ = 1 10 = 0.1 Find P ( x > 7). Draw the graph. P ( x > 7) = 1 – P ( x < 7). Since P ( X < x ) = 1 – e –mx then P ( X > x ) = 1 – ( 1 – e –mx ) = e –mx P ( x > 7) = e (–0.1)(7) = 0.4966. The probability that a computer part lasts more than seven years is 0.4966. b. On the average, how long would five computer parts last if they are used one after another? b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years. d. What is the probability that a computer part lasts between nine and 11 years? d. Find P (9 < x < 11). Draw the graph. P (9 < x < 11) = P ( x < 11) – P ( x < 9) = (1 – e (–0.1)(11) ) – (1 – e (–0.1)(9) ) = 0.6671 – 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737. Try It On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day? P ( x > 15) = 0.4346 Six pairs of running shoes would last 108 months on average. 80 th percentile = 28.97 months Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter 1 12 . The decay p[parameter is another way to view 1/λ. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes. What is m , μ , and σ ? The probability that you must wait more than five minutes is _______ . m = 1 12 μ = 12 σ = 12 P ( x > 5) = 0.6592 Try It Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter 1 20 . Let X = the distance people are willing to commute in miles. What is m , μ , and σ ? What is the probability that a person is willing to commute more than 25 miles? The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed. On average, how many minutes elapse between two successive arrivals? When the store first opens, how long on average does it take for three customers to arrive? After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive. Is an exponential distribution reasonable for this situation? Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive. Let X = the time between arrivals, in minutes. By part a, μ = 2, so m = 1 2 = 0.5. The cumulative distribution function is P ( X < x ) = 1 – e (-0.5)(x) Therefore P ( X < 1) = 1 – e (–0.5)(1) = 0.3935. P ( X > 5) = 1 – P ( X < 5) = 1 – (1 – e (-0.5)(5) ) = e –2.5 ≈ 0.0821. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others. Try It Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution. On average, how many seconds elapse between two successive cars? After a car passes by, how long on average will it take for another seven cars to pass by? Find the probability that after a car passes by, the next car will pass within the next 20 seconds. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds. Memorylessness of the Exponential Distribution Recall that the amount of time between customers for the postal clerk discussed earlier is exponentially distributed with a mean of two minutes. Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property . The exponential and geometric probability density functions are the only probability functions that have the memoryless property. Specifically, the memoryless property says that P ( X > r + t | X > r ) = P ( X > t ) for all r ≥ 0 and t ≥ 0 For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation. P ( X > 5 + 1 | X > 5) = P ( X > 1) = e ( – 0.5 ) ( 1 ) = 0.6065. This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival. The exponential distribution is often used to model the longevity of an electrical or mechanical device. In , the lifetime of a certain computer part has the exponential distribution with a mean of ten years. The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P ( X > 17| X > 10) = P ( X > 7) = 0.4966, where the vertical line is read as "given". Refer back to the postal clerk again where the time a postal clerk spends with a customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that the customer will spend at least an additional three minutes with the postal clerk? The decay parameter of X is m = 1 4 = 0.25, so X ∼ Exp (0.25). The cumulative distribution function is P ( X < x ) = 1 – e –0.25 x . We want to find P ( X > 7| X > 4). The memoryless property says that P ( X > 7| X > 4) = P ( X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk. This is P ( X > 3) = 1 – P ( X < 3) = 1 – (1 – e –0.25⋅3 ) = e –0.75 ≈ 0.4724. Try It Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years. Relationship between the Poisson and the Exponential Distribution There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean μ . Recall that if X has the Poisson distribution with mean μ , then P ( X = x ) = μ x e − μ x ! . The formula for the exponential distribution: P ( X = x ) = m e - m x = 1 μ e - 1 μ x Where m = the rate parameter, or μ = average time between occurrences. We see that the exponential is the cousin of the Poisson distribution and they are linked through this formula. There are important differences that make each distribution relevant for different types of probability problems. First, the Poisson has a discrete random variable, x, where time; a continuous variable is artificially broken into discrete pieces. We saw that the number of occurrences of an event in a given time interval, x, follows the Poisson distribution. For example, the number of times the telephone rings per hour. By contrast, the time between occurrences follows the exponential distribution. For example. The telephone just rang, how long will it be until it rings again? We are measuring length of time of the interval, a continuous random variable, exponential, not events during an interval, Poisson. The Exponential Distribution v. the Poisson Distribution A visual way to show both the similarities and differences between these two distributions is with a time line. The random variable for the Poisson distribution is discrete and thus counts events during a given time period, t 1 to t 2 on , and calculates the probability of that number occurring. The number of events, four in the graph, is measured in counting numbers; therefore, the random variable of the Poisson is a discrete random variable. The exponential probability distribution calculates probabilities of the passage of time, a continuous random variable. In this is shown as the bracket from t 1 to the next occurrence of the event marked with a triangle. Classic Poisson distribution questions are "how many people will arrive at my checkout window in the next hour?". Classic exponential distribution questions are "how long it will be until the next person arrives," or a variant, "how long will the person remain here once they have arrived?". Again, the formula for the exponential distribution is: f ( x ) = m e - m x or f ( x ) = 1 μ e - 1 μ x We see immediately the similarity between the exponential formula and the Poisson formula. P ( x ) = μ x e − μ x ! Both probability density functions are based upon the relationship between time and exponential growth or decay. The “e” in the formula is a constant with the approximate value of 2.71828 and is the base of the natural logarithmic exponential growth formula. When people say that something has grown exponentially this is what they are talking about. An example of the exponential and the Poisson will make clear the differences been the two. It will also show the interesting applications they have. Poisson Distribution Suppose that historically 10 customers arrive at the checkout lines each hour. Remember that this is still probability so we have to be told these historical values. We see this is a Poisson probability problem. We can put this information into the Poisson probability density function and get a general formula that will calculate the probability of any specific number of customers arriving in the next hour. The formula is for any value of the random variable we chose, and so the x is put into the formula. In this example, μ = 10 because we expect 10 customers in line each hour. This is the formula: f ( x ) = 10 x e -10 x ! As an example, the probability of 15 people arriving at the checkout counter in the next hour would be P ( x = 15 ) = 10 15 e -10 15 ! = 0.0347 Here we have inserted x = 15 and calculated the probability that in the next hour 15 people will arrive is 0.035. Exponential Distribution If we keep the same historical facts that 10 customers arrive each hour, but we now are interested in the service time a person spends at the counter, then we would use the exponential distribution. The exponential probability function for any value of x, the random variable, for this particular checkout counter historical data is: f ( x ) = 1 .1 e -x .1 = 10 e -10 x To calculate µ , the historical average service time, we simply divide the number of people that arrive per hour, 10 , into the time period, one hour, and have µ = 0.1. Historically, people spend 0.1 of an hour at the checkout counter, or 6 minutes. This explains the .1 in the formula. There is a natural confusion with µ in both the Poisson and exponential formulas. They have different meanings, although they have the same symbol. The mean of the exponential is one divided by the mean of the Poisson. If you are given the historical number of arrivals you have the mean of the Poisson. If you are given an historical length of time between events you have the mean of an exponential. Continuing with our example at the checkout clerk; if we wanted to know the probability that a person would spend 9 minutes or less checking out, then we use this formula. First, we convert to the same time units which are parts of one hour. Nine minutes is 0.15 of one hour. Next we note that we are asking for a range of values. This is always the case for a continuous random variable. We write the probability question as: p ( x ≤ 9 ) = 1 - 10 e -10 x We can now put the numbers into the formula and we have our result. p ( x = .15 ) = 1 - 10 e -10 ( .15 ) = 0.7769 The probability that a customer will spend 9 minutes or less checking out is 0.7769. We see that we have a high probability of getting out in less than nine minutes and a tiny probability of having 15 customers arriving in the next hour. At a police station in a large city, calls come in at an average rate of four calls per minute. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has a Poisson distribution. Find the average time between two successive calls. Find the probability that after a call is received, the next call occurs in less than ten seconds. Find the probability that exactly five calls occur within a minute. Find the probability that fewer than five calls occur within a minute. Find the probability that more than 40 calls occur in an eight-minute period. On average, four calls occur per minute, so one call occurs every 15 seconds, or = 0.25 minutes occur between successive calls on average. Let T = time elapsed between calls. From part a, μ = 0 . 25 , so m = 4 . Thus, T ∼ Exp (4). The cumulative distribution function is P ( T < t ) = 1 - e - 4 t ) . The probability that the next call occurs in less than ten seconds ( ten seconds = 1 6 minute ) is P T < 1 6 = 1 - e - 4 ( 1 6 ) = 0 . 4866 . Let X = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute. Therefore, X ~ P o i s s o n ( 4 ) and so P ( X = 5 ) = 4 5 e - 4 5 ! ≈ 0 . 1563 . 5 ! = ( 5 ) ( 4 ) ( 3 ) ( 2 ) ( 1 ) Keep in mind that X must be a whole number, so P ( X < 5 ) = P ( X ≤ 4 ) . To compute this, we could take P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) . Using technology, we see that P ( X ≤ 4 ) = 0 . 6288 . Let Y = the number of calls that occur during an eight-minute period. Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight-minute period. Hence, Y ~ P o i s s o n ( 32 ) . Therefore, P ( Y > 40 ) = 1 - P ( Y ≤ 40 ) = 1 - 0 . 9294 = 0 . 0707 . Try It In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week. Calculate the probability that there are at most 2 accidents occur in any given week. What is the probability that there is at least two weeks between any 2 accidents? Chapter Review If X has an exponential distribution with mean μ , then the decay parameter is m = 1 μ . The probability density function of X is f ( x ) = me -mx (or equivalently f ( x ) = 1 μ e − x / μ . The cumulative distribution function of X is P ( X ≤ x ) = 1 – e – mx . Formula Review pdf: f ( x ) = me (– mx ) where x ≥ 0 and m > 0 cdf: P ( X ≤ x ) = 1 – e (– mx ) mean µ = 1 m standard deviation σ = µ Additionally P ( X > x ) = e (– mx ) P ( a < X < b ) = e (– ma ) – e (– mb ) Poisson probability: P ( X = x ) = μ x e − μ x ! with mean and variance of μ References Data from the United States Census Bureau. Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013). “No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013). Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf (accessed June 11, 2013). Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp (0.2) What type of distribution is this? Are outcomes equally likely in this distribution? Why or why not? No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time. What is m ? What does it represent? What is the mean? five What is the standard deviation? State the probability density function. f ( x ) = 0.2e -0.2 x Graph the distribution. Find P (2 < x < 10). 0.5350 Find P ( x > 6). Find the 70 th percentile. 6.02 Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp (0.75). What is m ? What is the probability density function? f ( x ) = 0.75 e -0.75 x What is the cumulative distribution function? Draw the distribution. Find P ( x < 4). Find the 30 th percentile. 0.4756 Find the median. Which is larger, the mean or the median? The mean is larger. The mean is 1 m = 1 0.75 ≈ 1.33 , which is greater than 0.9242. Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14. What is being measured here? Are the data discrete or continuous? continuous In words, define the random variable X . What is the decay rate ( m )? m = 0.000121 The distribution for X is ______. Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P ( x < 5,730). Sketch the graph, and shade the area of interest. Find the probability. P ( x < 5,730) = __________ Answers may vary. P ( x < 5,730) = 0.5001 Find the percentage of carbon-14 lasting longer than 10,000 years. Sketch the graph, and shade the area of interest. Find the probability. P ( x > 10,000) = ________ Thirty percent (30%) of carbon-14 will decay within how many years? Sketch the graph, and shade the area of interest. Find the value k such that P ( x < k ) = 0.30. Answers may vary. k = 2947.73 Homework Suppose that the length of phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. Define the random variable. X = ________________. Is X continuous or discrete? μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that a phone call lasts less than nine minutes. Find the probability that a phone call lasts more than nine minutes. Find the probability that a phone call lasts between seven and nine minutes. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why? Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery. Define the random variable. X = _________________________________. Is X continuous or discrete? On average, how long would you expect one car battery to last? On average, how long would you expect nine car batteries to last, if they are used one after another? Find the probability that a car battery lasts more than 36 months. Seventy percent of the batteries last at least how long? X = the useful life of a particular car battery, measured in months. X is continuous. 40 months 360 months 0.4066 14.27 At one point in time, the percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state. Define the random variable. X = _________________________________. Is X continuous or discrete? μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that the percent is less than 12. Find the probability that the percent is between eight and 14. The percent of all individuals living in the United States who speak a language at home other than English is 13.8. Why is this number different from 9.848%? What would make this number higher than 9.848%? The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement. Define the random variable. X = _________________________________. Is X continuous or discrete? μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that the person retired after age 70. Do more people retire before age 65 or after age 65? In a room of 1,000 people over age 80, how many do you expect will NOT have retired yet? X = the time (in years) after reaching age 60 that it takes an individual to retire X is continuous. five five Answers may vary. 0.1353 before 18.3 The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150. Define the random variable. X = _________________________________. μ = ________ σ = ________ Draw a graph of the probability distribution. Label the axes. Find the probability that a car required over $300 for maintenance during its first year. Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution. The decay rate is: 0.3333 0.5000 2 3 a What is the probability that a smartphones will fail within two years of the date of purchase? 0.8647 0.4866 0.2212 0.9997 What is the median lifetime of these smartphones (in years)? 0.1941 1.3863 2.0794 5.5452 c At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution. On average, how much time occurs between five consecutive calls? Find the probability that after a call is received, it takes more than three minutes for the next call to occur. Ninety-percent of all calls occur within how many minutes of the previous call? Suppose that two minutes have elapsed since the last call. Find the probability that the next call will occur within the next minute. Find the probability that less than 20 calls occur within an hour. In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential. What is the probability that an entire season elapses with a single no-hitter? If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season? What is the probability that there are more than 3 no-hitters in a single season? Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3. Therefore, ( X = 0) = 3 0 e – 3 0 ! = e –3 ≈ 0.0498 NOTE You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is 1 3 season. For the exponential, µ = 1 3 . Therefore, m = 1 μ = 3 and T ∼ Exp (3). The desired probability is P ( T > 1) = 1 – P ( T < 1) = 1 – (1 – e –3 ) = e –3 ≈ 0.0498. Let T = duration of time between no-hitters. We find P ( T > 2| T > 1), and by the memoryless property this is simply P ( T > 1), which we found to be 0.0498 in part a. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P ( X > 3) = 1 – P ( X ≤ 3) = 0.3528. During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential. What is the probability that the next earthquake occurs within the next three months? Given that six months has passed without an earthquake in Papua New Guinea, what is the probability that the next three months will be free of earthquakes? What is the probability of zero earthquakes occurring in 2014? What is the probability that at least two earthquakes will occur in 2014? According to the American Red Cross, about one out of nine people in the U.S. have Type B positive blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B positive blood types that arrive roughly follows the Poisson distribution. If 100 people arrive, how many on average would be expected to have Type B positive blood? What is the probability that over 10 people out of these 100 have type B positive blood? What is the probability that more than 20 people arrive before a person with type B positive blood is found? 100 9 = 11.11 P ( X > 10) = 1 – P ( X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532. The number of people with Type B positive blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9 . The cumulative distribution function of X is P ( X < x ) = 1 − e − x 9 . Thus, P ( X > 20) = 1 - P ( X ≤ 20) = 1 − ( 1 − e − 20 9 ) ≈ 0.1084. NOTE We could also deduce that each person arriving has a 8/9 chance of not having Type B positive blood. So the probability that none of the first 20 people arrive have Type B positive blood is ( 8 9 ) 20 ≈ 0.0948 . (The geometric distribution is more appropriate than the exponential because the number of people between Type B positive people is discrete instead of continuous.) A web site experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution. Find the probability that the duration between two successive visits to the web site is more than ten minutes. The top 25% of durations between visits are at least how long? Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes? Find the probability that less than 7 visits occur within a one-hour period. At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed. Find the probability that the time between two successive visits to the urgent care facility is less than 2 minutes. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes? Find the probability that more than eight patients arrive during a half-hour period. Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7 . The cdf is P ( T < t ) = 1 − e t 7 P ( T < 2) = 1 - 1 − e − 2 7 ≈ 0.2485. P ( T > 15) = 1 − P ( T < 15 ) = 1 − ( 1 − e − 15 7 ) ≈ e − 15 7 ≈ 0.1173 . P ( T > 15| T > 10) = P ( T > 5) = 1 − ( 1 − e − 5 7 ) = e − 5 7 ≈ 0.4895 . Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7 , X ∼ Poisson ( 30 7 ) . Find P ( X > 8) = 1 – P ( X ≤ 8) ≈ 0.0311. decay parameter The decay parameter describes the rate at which probabilities decay to zero for increasing values of x . It is the value m in the probability density function f ( x ) = me (- mx ) of an exponential random variable. It is also equal to m = 1 μ , where μ is the mean of the random variable. memoryless property For an exponential random variable X , the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that X exceeds x + t , given that it has exceeded x , is the same as the probability that X would exceed t if we had no knowledge about it. In symbols we say that P ( X > x + t | X > x ) = P ( X > t ). Poisson distribution If there is a known average of μ events occurring per unit time, and these events are independent of each other, then the number of events X occurring in one unit of time has the Poisson distribution. The probability of x events occurring in one unit time is equal to P ( X = x ) = μ x e − μ x ! .

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Introduction If you ask enough people about their shoe size, you will find that your graphed data is shaped like a bell curve and can be described as normally distributed. (credit: modification of work “Shoe shop” by Lo/ Flickr, Public Domain) The normal probability density function, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines. Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world. Remember here that we are still talking about the distribution of population data. This is a discussion of probability and thus it is the population data that may be normally distributed, and if it is, then this is how we can find probabilities of specific events just as we did for population data that may be binomially distributed or Poisson distributed. This caution is here because in the next chapter we will see that the normal distribution describes something very different from raw data and forms the foundation of inferential statistics. The normal distribution has two parameters (two numerical descriptive measures): the mean ( μ ) and the standard deviation ( σ ). If X is a quantity to be measured that has a normal distribution with mean ( μ ) and standard deviation ( σ ), we designate this by writing the following formula of the normal probability density function: The probability density function is a rather complicated mathematical function. Do not memorize it . It is not necessary. f ( x ) = 1 σ ⋅ 2 ⋅ π ⋅ e − 1 2 ⋅ ( x − μ σ ) 2 The curve is symmetric about a vertical line drawn through the mean, μ . The mean is the same as the median, which is the same as the mode, because the graph is symmetric about μ . As the notation indicates, the normal distribution depends only on the mean and the standard deviation. Note that this is unlike several probability density functions we have already studied, such as the Poisson, where the mean is equal to μ and the standard deviation simply the square root of the mean, or the binomial, where p is used to determine both the mean and standard deviation. Since the area under the curve must equal one, a change in the standard deviation, σ , causes a change in the shape of the normal curve; the curve becomes fatter and wider or skinnier and taller depending on σ . A change in μ causes the graph to shift to the left or right. This means there are an infinite number of normal probability distributions. One of special interest is called the standard normal distribution . Formula Review X ∼ N ( μ , σ ) μ = the mean σ = the standard deviation Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 σ 2 2 , where μ is the mean of the distribution and σ is the standard deviation; notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV, Z, is called the standard normal distribution .

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The Standard Normal Distribution The standard normal distribution is a normal distribution of standardized values called z -scores . A z -score is measured in units of the standard deviation. The mean for the standard normal distribution is zero, and the standard deviation is one. What this does is dramatically simplify the mathematical calculation of probabilities. Take a moment and substitute zero and one in the appropriate places in the above formula and you can see that the equation collapses into one that can be much more easily solved using integral calculus. The transformation z = x − μ σ produces the distribution Z ~ N (0, 1). The value x in the given equation comes from a known normal distribution with known mean μ and known standard deviation σ . The z -score tells how many standard deviations a particular x is away from the mean. Z -Scores If X is a normally distributed random variable and X ~ N(μ, σ) , then the z -score for a particular x is: z = x – μ σ The z -score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ . Values of x that are larger than the mean have positive z -scores, and values of x that are smaller than the mean have negative z -scores. If x equals the mean, then x has a z -score of zero. Suppose X ~ N(5, 6) . This says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then: z = x – μ σ = 17 – 5 6 = 2 This means that x = 17 is two standard deviations (2 σ ) above or to the right of the mean μ = 5. Now suppose x = 1. Then: z = x – μ σ = 1 – 5 6 = –0.67 (rounded to two decimal places) This means that x = 1 is 0.67 standard deviations (–0.67 σ ) below or to the left of the mean μ = 5. Try It Fill in the blanks. Jerome averages 16 points a game with a standard deviation of four points. X ~ N (16,4). Suppose Jerome scores ten points in a game. The z –score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?). The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ , then the Empirical Rule states the following: About 68% of the x values lie between –1 σ and +1 σ of the mean µ (within one standard deviation of the mean). About 95% of the x values lie between –2 σ and +2 σ of the mean µ (within two standard deviations of the mean). About 99.7% of the x values lie between –3 σ and +3 σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean. The z -scores for +1 σ and –1 σ are +1 and –1, respectively. The z -scores for +2 σ and –2 σ are +2 and –2, respectively. The z -scores for +3 σ and –3 σ are +3 and –3 respectively. Suppose x has a normal distribution with mean 50 and standard deviation 6. About 68% of the x values lie within one standard deviation of the mean. Therefore, about 68% of the x values lie between –1 σ = (–1)(6) = –6 and 1 σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z -scores are –1 and +1 for 44 and 56, respectively. About 95% of the x values lie within two standard deviations of the mean. Therefore, about 95% of the x values lie between –2 σ = (–2)(6) = –12 and 2 σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z -scores are –2 and +2 for 38 and 62, respectively. About 99.7% of the x values lie within three standard deviations of the mean. Therefore, about 99.7% of the x values lie between –3 σ = (–3)(6) = –18 and 3 σ = (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations from the mean 50. The z -scores are –3 and +3 for 32 and 68, respectively. Try It Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68% of the values lie? References “Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013). “The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013). “2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013). “Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013). Data from the San Jose Mercury News . Data from The World Almanac and Book of Facts . “List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013). Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013). Chapter Review A z -score is a standardized value. Its distribution is the standard normal, Z ~ N (0, 1). The mean of the z -scores is zero and the standard deviation is one. If z is the z -score for a value x from the normal distribution N ( µ , σ ) then z tells you how many standard deviations x is above (greater than) or below (less than) µ . Formula Review Z ~ N (0, 1) z = a standardized value ( z -score) mean = 0; standard deviation = 1 To find the observed value, x , when the z -scores is known: x = μ + ( z ) σ z -score: z = x – μ σ or z = | x – μ | σ Z = the random variable for z -scores Z ~ N (0, 1) A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________. ounces of water in a bottle A normal distribution has a mean of 61 and a standard deviation of 15. What is the median? X ~ N (1, 2) σ = _______ 2 A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________. X ~ N (–4, 1) What is the median? –4 X ~ N (3, 5) σ = _______ X ~ N (–2, 1) μ = _______ –2 What does a z -score measure? What does standardizing a normal distribution do to the mean? The mean becomes zero. Is X ~ N (0, 1) a standardized normal distribution? Why or why not? What is the z -score of x = 12, if it is two standard deviations to the right of the mean? z = 2 What is the z -score of x = 9, if it is 1.5 standard deviations to the left of the mean? What is the z -score of x = –2, if it is 2.78 standard deviations to the right of the mean? z = 2.78 What is the z -score of x = 7, if it is 0.133 standard deviations to the left of the mean? Suppose X ~ N (2, 6). What value of x has a z -score of three? x = 20 Suppose X ~ N (8, 1). What value of x has a z -score of –2.25? Suppose X ~ N (9, 5). What value of x has a z -score of –0.5? x = 6.5 Suppose X ~ N (2, 3). What value of x has a z -score of –0.67? Suppose X ~ N (4, 2). What value of x is 1.5 standard deviations to the left of the mean? x = 1 Suppose X ~ N (4, 2). What value of x is two standard deviations to the right of the mean? Suppose X ~ N (8, 9). What value of x is 0.67 standard deviations to the left of the mean? x = 1.97 Suppose X ~ N (–1, 2). What is the z -score of x = 2? Suppose X ~ N (12, 6). What is the z -score of x = 2? z = –1.67 Suppose X ~ N (9, 3). What is the z -score of x = 9? Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z -score of x = 5.5? z ≈ –0.33 In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean. 0.67, right In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean. In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean. 3.14, left In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean. About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution? about 68% About what percent of the x values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution? About what percent of x values lie between the second and third standard deviations (both sides)? about 4% Suppose X ~ N (15, 3). Between what x values does 68.27% of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15). Suppose X ~ N (–3, 1). Between what x values does 95.45% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3). between –5 and –1 Suppose X ~ N (–3, 1). Between what x values does 34.14% of the data lie? About what percent of x values lie between the mean and three standard deviations? about 50% About what percent of x values lie between the mean and one standard deviation? About what percent of x values lie between the first and second standard deviations from the mean (both sides)? about 27% About what percent of x values lie between the first and third standard deviations(both sides)? Use the following information to answer the next two exercises: The life of wearable fitness devicess is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Define the random variable X in words. X = _______________. The lifetime of a wearable fitness device measured in years. X ~ _____(_____,_____) Homework Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the median recovery time? 2.7 5.3 7.4 2.1 What is the z -score for a patient who takes ten days to recover? 1.5 0.2 2.2 7.3 c The length of time to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true? The data cannot follow the uniform distribution. The data cannot follow the exponential distribution.. The data cannot follow the normal distribution. I only II only III only I, II, and III The heights of the 430 National Basketball Association players were listed on team rosters at the start of a recent season. The heights of basketball players have an approximate normal distribution with mean, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z -score and interpret it using complete sentences. 77 inches 85 inches If an NBA player reported his height had a z -score of 3.5, would you believe him? Explain your answer. Use the z -score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. Use the z -score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely. The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution. Calculate the z -scores for the male systolic blood pressures 100 and 150 millimeters. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, but that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him? Kyle’s doctor told him that the z -score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score then X ~ N (125, 14). Which answer(s) is/are correct? Kyle’s systolic blood pressure is 175. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age. Kyle’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. Calculate Kyle’s blood pressure. iv Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5. Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N (10.2, 0.8). Calculate the z -scores that correspond to the following weights and interpret them. 11 kg 7.9 kg 12.2 kg During a certain year, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115. Calculate the z -score for an SAT score of 720. Interpret it using a complete sentence. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? During a different year, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took? Let X = an SAT math score and Y = an ACT math score. X = 720 : z = 720 - 520 115 = 1 . 74 . The exam score of 720 is 1.74 standard deviations above the mean of 520. z = 1.5 The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. X – μ σ = 700 – 514 117 ≈ 1.59, the z -score for the SAT. Y – μ σ = 30 – 21 5.3 ≈ 1.70, the z -scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z -score). Standard Normal Distribution a continuous random variable (RV) X ~ N (0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N (0, 1). z-score the linear transformation of the form z = x - μ σ or written as z = | x – μ | σ ; if this transformation is applied to any normal distribution X ~ N ( μ , σ ) the result is the standard normal distribution Z ~ N (0,1). If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ , the result is called the z -score of x . The z -score allows us to compare data that are normally distributed but scaled differently. A z -score is the number of standard deviations a particular x is away from its mean value.

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Using the Normal Distribution The shaded area in the following graph indicates the area to the right of x . This area is represented by the probability P ( X > x ). Normal tables provide the probability between the mean, zero for the standard normal distribution, and a specific value such as x 1 . This is the unshaded part of the graph from the mean to x 1 . Because the normal distribution is symmetrical , if x 1 were the same distance to the left of the mean the area, probability, in the left tail, would be the same as the shaded area in the right tail. Also, bear in mind that because of the symmetry of this distribution, one-half of the probability is to the right of the mean and one-half is to the left of the mean. Calculations of Probabilities To find the probability for probability density functions with a continuous random variable we need to calculate the area under the function across the values of X we are interested in. For the normal distribution this seems a difficult task given the complexity of the formula. There is, however, a simply way to get what we want. Here again is the formula for the normal distribution: f ( x ) = 1 σ ⋅ 2 ⋅ π ⋅ e − 1 2 ⋅ ( x − μ σ ) 2 Looking at the formula for the normal distribution it is not clear just how we are going to solve for the probability doing it the same way we did it with the previous probability functions. There we put the data into the formula and did the math. To solve this puzzle we start knowing that the area under a probability density function is the probability. This shows that the area between X 1 and X 2 is the probability as stated in the formula: P (X 1 ≤ x ≤ X 2 ) The mathematical tool needed to find the area under a curve is integral calculus. The integral of the normal probability density function between the two points x 1 and x 2 is the area under the curve between these two points and is the probability between these two points. Doing these integrals is no fun and can be very time consuming. But now, remembering that there are an infinite number of normal distributions out there, we can consider the one with a mean of zero and a standard deviation of 1. This particular normal distribution is given the name Standard Normal Distribution. Putting these values into the formula it reduces to a very simple equation. We can now quite easily calculate all probabilities for any value of x, for this particular normal distribution, that has a mean of zero and a standard deviation of 1. These have been produced and are available here in the appendix to the text or everywhere on the web. They are presented in various ways. The table in this text is the most common presentation and is set up with probabilities for one-half the distribution beginning with zero, the mean, and moving outward. The shaded area in the graph at the top of the table in Statistical Tables represents the probability from zero to the specific Z value noted on the horizontal axis, Z. The only problem is that even with this table, it would be a ridiculous coincidence that our data had a mean of zero and a standard deviation of one. The solution is to convert the distribution we have with its mean and standard deviation to this new Standard Normal Distribution. The Standard Normal has a random variable called Z. Using the standard normal table, typically called the normal table, to find the probability of one standard deviation, go to the Z column, reading down to 1.0 and then read at column 0. That number, 0.3413 is the probability from zero to 1 standard deviation. At the top of the table is the shaded area in the distribution which is the probability for one standard deviation. The table has solved our integral calculus problem. But only if our data has a mean of zero and a standard deviation of 1. However, the essential point here is, the probability for one standard deviation on one normal distribution is the same on every normal distribution. If the population data set has a mean of 10 and a standard deviation of 5 then the probability from 10 to 15, one standard deviation, is the same as from zero to 1, one standard deviation on the standard normal distribution. To compute probabilities, areas, for any normal distribution, we need only to convert the particular normal distribution to the standard normal distribution and look up the answer in the tables. As review, here again is the standardizing formula : Z = x - μ σ where Z is the value on the standard normal distribution, X is the value from a normal distribution one wishes to convert to the standard normal, μ and σ are, respectively, the mean and standard deviation of that population. Note that the equation uses μ and σ which denotes population parameters. This is still dealing with probability so we always are dealing with the population, with known parameter values and a known distribution. It is also important to note that because the normal distribution is symmetrical it does not matter if the z-score is positive or negative when calculating a probability. One standard deviation to the left (negative Z-score) covers the same area as one standard deviation to the right (positive Z-score). This fact is why the Standard Normal tables do not provide areas for the left side of the distribution. Because of this symmetry, the Z-score formula is sometimes written as: Z = | x - μ | σ Where the vertical lines in the equation means the absolute value of the number. What the standardizing formula is really doing is computing the number of standard deviations X is from the mean of its own distribution. The standardizing formula and the concept of counting standard deviations from the mean is the secret of all that we will do in this statistics class. The reason this is true is that all of statistics boils down to variation, and the counting of standard deviations is a measure of variation. This formula, in many disguises, will reappear over and over throughout this course. The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. a. Find the probability that a randomly selected student scored more than 65 on the exam. b. Find the probability that a randomly selected student scored less than 85. a. Let X = a score on the final exam. X ~ N (63, 5), where μ = 63 and σ = 5. Draw a graph. Then, find P ( x > 65). P ( x > 65) = 0.3446 Z 1 = x 1 − μ σ = 65 − 63 5 = 0.4 P ( x ≥ x 1 ) = P ( Z ≥ Z 1 ) = 0.3446 The probability that any student selected at random scores more than 65 is 0.3446. Here is how we found this answer. The normal table provides probabilities from zero to the value Z 1 . For this problem the question can be written as: P(X ≥ 65) = P(Z ≥ Z 1 ), which is the area in the tail. To find this area the formula would be 0.5 – P(X ≤ 65). One half of the probability is above the mean value because this is a symmetrical distribution. The graph shows how to find the area in the tail by subtracting that portion from the mean, zero, to the Z 1 value. The final answer is: P(X ≥ 63) = P(Z ≥ 0.4) = 0.3446 z = 65 – 63 5 = 0.4 Area to the left of Z 1 to the mean of zero is 0.1554 P ( x > 65) = P ( z > 0.4) = 0.5 – 0.1554 = 0.3446 b. Z = x - μ σ = 85 - 63 5 = 4.4 which is larger than the maximum value on the Standard Normal Table. Therefore, the probability that one student scores less than 85 is approximately one or 100%. A score of 85 is 4.4 standard deviations from the mean of 63 which is beyond the range of the standard normal table. Therefore, the probability that one student scores less than 85 is approximately one (or 100%). Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65. normalcdf(0,65,68,3) = 0.1587 A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day. a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N (2, 0.5) where μ = 2 and σ = 0.5. Find P (1.8 < x < 2.75). The probability for which you are looking is the area between x = 1.8 and x = 2.75. P (1.8 < x < 2.75) = 0.5886 P (1.8 ≤ x ≤ 2.75) = P ( Z i ≤ Z ≤ Z 2 ) The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886. b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment. b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25 th percentile, k , where P ( x < k ) = 0.25. f ( Z ) = 0.5 - 0.25 = 0.25 , therefore Z ≈ -0.675 ( or just 0.67 using the table ) Z = x - μ σ = x - 2 0.5 = -0.675 , therefore x = -0.675 * 0.5 + 2 = 1.66 hours. The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours. Try It The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70. normalcdf(66,70,68,3) = 0.4950 In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively. a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old. a. 0.8186 b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old. b. 0.8413 Try It Use the information in to answer the following questions. Find the 30 th percentile, and interpret it in a complete sentence. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old. A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on their farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph. Z 1 = 6 − 5.85 .24 = .625 P ( x ≥ 6) = P ( z ≥ 0.625) = 0.2670 b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______. f ( Z ) = 0.20 2 = 0.10 , therefore Z ≈ ± 0.25 Z = x - μ σ = x - 5.85 0.24 = ± 0.25 → ± 0.25 · 0.24 + 5.85 = ( 5.79 , 5.91 ) Try It Using the information from , answer the following: The middle 40% of mandarin oranges from this farm are between ______ and ______. Find the 16 th percentile and interpret it in a complete sentence. References “Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013). “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013). “Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013). “Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013). “Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebook-statistics/(accessed May 14, 2013).

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