diff --git "a/math-ds-complete/index.html" "b/math-ds-complete/index.html" --- "a/math-ds-complete/index.html" +++ "b/math-ds-complete/index.html" @@ -810,6 +810,90 @@

For skewed data (like income, house prices), always report the median along with the mean. If they're very different, your data has outliers or is skewed!

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Find the mean, median, and mode of: [12, 15, 12, 18, 20, 15, 12, 22]

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Calculate the Mean (Average)

+
+ Sum = 12 + 15 + 12 + 18 + 20 + 15 + 12 + 22 = 126
+ Count (n) = 8 values
+ Mean = Sum ÷ n = 126 ÷ 8 = 15.75 +
+

Add all values together, then divide by how many values there are

+
+
+ +
+
Step 2:
+
+

Find the Median (Middle Value)

+
+ Sorted data: [12, 12, 12, 15, 15, 18, 20, 22]
+ Even number of values (8), so average the middle two
+ Middle positions: 4th and 5th values = 15 and 15
+ Median = (15 + 15) ÷ 2 = 15 +
+

For even-sized datasets, average the two middle values

+
+
+ +
+
Step 3:
+
+

Find the Mode (Most Frequent Value)

+
+ Frequency count:
+ • 12 appears 3 times ← Most frequent!
+ • 15 appears 2 times
+ • 18, 20, 22 each appear 1 time
+ Mode = 12 +
+

The mode is the value that appears most often

+
+
+ +
+ Final Answer: + Mean = 15.75, Median = 15, Mode = 12 +
+ +
+ ✓ Check: +

Mean (15.75) is slightly higher than median (15) because the outlier 22 pulls it up. The mode (12) is the lowest because it's the most common value at the lower end.

+
+
+ +
+

💪 Try These:

+
    +
  1. Find the mean of: [5, 10, 15, 20]
    2. +
  2. What's the median of: [3, 1, 4, 1, 5]?
    3. +
  3. Find the mode of: [2, 2, 3, 4, 4, 4, 5]
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Calculate the variance and standard deviation for the dataset: [4, 8, 6, 5, 3]

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Calculate the Mean

+
+ Sum = 4 + 8 + 6 + 5 + 3 = 26
+ Mean (x̄) = 26 ÷ 5 = 5.2 +
+

First, we need the mean to calculate deviations

+
+
+ +
+
Step 2:
+
+

Find Deviations from Mean

+
+ (4 - 5.2) = -1.2
+ (8 - 5.2) = 2.8
+ (6 - 5.2) = 0.8
+ (5 - 5.2) = -0.2
+ (3 - 5.2) = -2.2 +
+

Subtract the mean from each value

+
+
+ +
+
Step 3:
+
+

Square Each Deviation

+
+ (-1.2)² = 1.44
+ (2.8)² = 7.84
+ (0.8)² = 0.64
+ (-0.2)² = 0.04
+ (-2.2)² = 4.84 +
+

Squaring eliminates negative signs and emphasizes larger deviations

+
+
+ +
+
Step 4:
+
+

Calculate Variance (sample)

+
+ Sum of squared deviations = 1.44 + 7.84 + 0.64 + 0.04 + 4.84 = 14.8
+ Divide by (n-1) = 5-1 = 4
+ s² = 14.8 ÷ 4 = 3.7 +
+

We use (n-1) for sample variance (Bessel's correction)

+
+
+ +
+
Step 5:
+
+

Calculate Standard Deviation

+
+ s = √s² = √3.7 ≈ 1.92 +
+

Standard deviation is the square root of variance

+
+
+ +
+ Final Answer: + Variance = 3.7, Standard Deviation = 1.92 +
+ +
+ ✓ Interpretation: +

A standard deviation of 1.92 means most values fall within about 1.92 units of the mean (5.2). This indicates moderate spread in the data.

+
+
+ +
+

💪 Try These:

+
    +
  1. Calculate the standard deviation of: [2, 4, 6, 8]
    2. +
  2. Find the variance of: [10, 12, 14, 16, 18]
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Calculate and interpret skewness for dataset: [2, 3, 4, 5, 15]

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Calculate the Mean

+
+ Sum = 2 + 3 + 4 + 5 + 15 = 29
+ n = 5
+ Mean (x̄) = 29/5 = 5.8 +
+

First, find the average of all values

+
+
+ +
+
Step 2:
+
+

Calculate Standard Deviation

+
+ Deviations from mean: (2-5.8), (3-5.8), (4-5.8), (5-5.8), (15-5.8)
+ = -3.8, -2.8, -1.8, -0.8, 9.2
+ Squared: 14.44, 7.84, 3.24, 0.64, 84.64
+ Variance (sample) = (14.44+7.84+3.24+0.64+84.64)/4 = 110.8/4 = 27.7
+ SD = √27.7 = 5.26 +
+

We need standard deviation for the skewness formula

+
+
+ +
+
Step 3:
+
+

Calculate Skewness

+
+ Cubed deviations: (-3.8)³, (-2.8)³, (-1.8)³, (-0.8)³, (9.2)³
+ = -54.87, -21.95, -5.83, -0.51, 778.69
+ Sum = 695.53
+ Skewness = (695.53/5) / (5.26)³ = 139.11 / 145.77 = 0.95 +
+

Skewness formula uses cubed deviations divided by cubed standard deviation

+
+
+ +
+
Step 4:
+
+

Interpret the Result

+
+ Skewness = +0.95 (positive)
+ Distribution is right-skewed
+ The value 15 pulls the tail to the right
+ Most data clustered on left, long tail on right +
+

Positive skewness means tail extends to the right

+
+
+ +
+ ✓ Final Answer: + Skewness = +0.95 (positively skewed, right tail) +
+ +
+ Check: +

The positive skewness confirms that the outlier (15) creates a long right tail, pulling the mean (5.8) above the median (4).

+
+
+ +
+

💪 Try These:

+
    +
  1. Find skewness of [1, 1, 2, 3, 3]
    2. +
  2. Data with left tail - positive or negative skew?
    3. +
  3. If mean < median, what type of skew?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Find covariance between X=[2, 4, 6, 8] and Y=[1, 3, 5, 7]

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Calculate the Means

+
+ x̄ = (2 + 4 + 6 + 8) / 4 = 20 / 4 = 5
+ ȳ = (1 + 3 + 5 + 7) / 4 = 16 / 4 = 4 +
+

Find the average of each variable

+
+
+ +
+
Step 2:
+
+

Create Deviation Table

+
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
xy(x-x̄)(y-ȳ)(x-x̄)(y-ȳ)
21-3-39
43-1-11
65111
87339
Sum20
+
+

Calculate deviations from means and their products

+
+
+ +
+
Step 3:
+
+

Calculate Sample Covariance

+
+ Cov(X,Y) = Σ(x-x̄)(y-ȳ) / (n-1)
+ Cov(X,Y) = 20 / (4-1)
+ Cov(X,Y) = 20 / 3
+ Cov(X,Y) = 6.67 +
+

Use n-1 for sample covariance (Bessel's correction)

+
+
+ +
+
Step 4:
+
+

Interpret the Result

+
+ Cov(X,Y) = 6.67 > 0
+ Positive covariance indicates:
+ • X and Y tend to increase together
+ • When X is above its mean, Y tends to be above its mean
+ • When X is below its mean, Y tends to be below its mean +
+

Positive covariance shows positive relationship

+
+
+ +
+ Final Answer: + Cov(X,Y) = 6.67 (positive relationship) +
+ +
+ ✓ Verification: +

The positive covariance confirms that X and Y have a positive linear relationship. As X increases by 2, Y also increases by 2, showing consistent movement together.

+
+
+ +
+

💪 Try These:

+
    +
  1. Calculate Cov(X,Y) for X=[1, 2, 3] and Y=[2, 4, 6]
    2. +
  2. If Cov(X,Y) = -5, what does this tell you about the relationship?
    3. +
  3. Find Cov(X,Y) for X=[5, 5, 5] and Y=[1, 2, 3]. What do you notice?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Calculate correlation coefficient for X=[2, 4, 6, 8] and Y=[1, 3, 5, 7]

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Use Covariance from Topic 11

+
+ From previous calculation:
+ Cov(X,Y) = 6.67
+ x̄ = 5, ȳ = 4 +
+

We already calculated this in Topic 11

+
+
+ +
+
Step 2:
+
+

Calculate Standard Deviation of X

+
+ Deviations from mean: -3, -1, 1, 3
+ Squared deviations: 9, 1, 1, 9
+ Sum of squared deviations = 20
+ Variance_x = 20 / (4-1) = 20/3 = 6.67
+ SD_x = √6.67 ≈ 2.58 +
+

Standard deviation measures spread of X values

+
+
+ +
+
Step 3:
+
+

Calculate Standard Deviation of Y

+
+ Deviations from mean: -3, -1, 1, 3
+ Squared deviations: 9, 1, 1, 9
+ Sum of squared deviations = 20
+ Variance_y = 20 / (4-1) = 20/3 = 6.67
+ SD_y = √6.67 ≈ 2.58 +
+

Standard deviation measures spread of Y values

+
+
+ +
+
Step 4:
+
+

Calculate Correlation Coefficient

+
+ r = Cov(X,Y) / (SD_x × SD_y)
+ r = 6.67 / (2.58 × 2.58)
+ r = 6.67 / 6.66
+ r ≈ 1.00 +
+

Correlation standardizes covariance by dividing by both standard deviations

+
+
+ +
+
Step 5:
+
+

Interpret the Result

+
+ r = 1.00 (perfect positive correlation)
+ This means:
+ • X and Y have a perfect linear relationship
+ • As X increases by 2, Y increases by 2 (exactly)
+ • All points lie exactly on a straight line
+ • The relationship is: Y = 0.5X (or Y = -1 + 0.5X when adjusted) +
+

r = 1 indicates perfect positive linear correlation

+
+
+ +
+ Final Answer: + r = 1.00 (perfect positive linear correlation) +
+ +
+ ✓ Verification: +

Check: If we plot these points, they form a perfect line. When X=2, Y=1; X=4, Y=3; X=6, Y=5; X=8, Y=7. The relationship is Y = (X/2) - 1 + (X/2) = 0.5X, which is indeed perfectly linear! ✓

+
+
+ +
+

💪 Try These:

+
    +
  1. If Cov(X,Y) = 10, SD_x = 2, SD_y = 5, find r
    2. +
  2. What does r = -0.8 indicate about the relationship?
    3. +
  3. Can correlation be greater than 1? Why or why not?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Study finds r = -0.75 between hours of TV watched and exam scores. Interpret this result and discuss causation.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Analyze the Sign

+
+ Negative correlation (r < 0)
+ As one variable increases, the other decreases
+ More TV → Lower scores (or vice versa) +
+

The negative sign tells us the direction of the relationship

+
+
+ +
+
Step 2:
+
+

Analyze the Strength

+
+ |r| = |-0.75| = 0.75
+ Interpretation scale:
+ • 0.0-0.3 = Weak
+ • 0.3-0.7 = Moderate
+ • 0.7-1.0 = Strong
+ 0.75 falls in "Strong" category +
+

The absolute value determines relationship strength

+
+
+ +
+
Step 3:
+
+

State the Relationship

+
+ Strong negative correlation
+ Students who watch more TV tend to have lower exam scores
+ Relationship is fairly consistent but not perfect +
+

Combine sign and strength for complete interpretation

+
+
+ +
+
Step 4:
+
+

Address Causation

+
+ Correlation ≠ Causation!
+ Possible explanations:
+ a) TV causes lower scores (less study time)
+ b) Lower-performing students watch more TV (compensating)
+ c) Third variable: stress causes both TV watching and poor performance
+ Cannot determine causation from correlation alone +
+

Correlation never proves causation - always consider alternatives

+
+
+ +
+
Step 5:
+
+

Predict Using Correlation

+
+ If we know TV hours, we can predict exam score
+ But prediction ≠ causation
+ r² = 0.75² = 0.56 = 56% of variance explained +
+

r² shows percentage of variance in one variable explained by the other

+
+
+ +
+ ✓ Final Answer: + Strong negative correlation (r = -0.75), but does NOT prove TV causes lower scores +
+ +
+ Check: +

While the correlation is strong, we must resist concluding causation. The relationship could be coincidental, reverse-causal, or due to confounding variables.

+
+
+ +
+

💪 Try These:

+
    +
  1. r = +0.90 between study hours and grades. Interpret.
    2. +
  2. Can r = 1.5? Why or why not?
    3. +
  3. If r = 0, does that mean no relationship at all?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

In a class of 40 students: 25 like Math, 20 like Science, 10 like both. Find: a) P(Math OR Science), b) P(only Math), c) P(neither)

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Set Up the Information

+
+ Total students: n = 40
+ P(Math) = 25/40 = 0.625
+ P(Science) = 20/40 = 0.5
+ P(Math ∩ Science) = 10/40 = 0.25 +
+

Convert all counts to probabilities

+
+
+ +
+
Step 2:
+
+

Find P(Math ∪ Science) using Addition Rule

+
+ Formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
+ P(Math ∪ Science) = 0.625 + 0.5 - 0.25
+ = 1.125 - 0.25
+ = 0.875 +
+

We subtract the intersection to avoid double-counting

+
+
+ +
+
Step 3:
+
+

Find P(only Math)

+
+ Only Math = Math AND NOT Science
+ Students in only Math = 25 - 10 = 15
+ P(only Math) = 15/40 = 0.375 +
+

Subtract those who like both from total Math students

+
+
+ +
+
Step 4:
+
+

Find P(neither)

+
+ Neither = NOT (Math OR Science)
+ P(neither) = 1 - P(Math ∪ Science)
+ = 1 - 0.875
+ = 0.125
+ Or: 40 - 35 = 5 students, so 5/40 = 0.125 ✓ +
+

Use complement rule or count directly

+
+
+ +
+ ✓ Final Answer: + a) P(Math OR Science) = 0.875 (87.5%)
b) P(only Math) = 0.375 (37.5%)
c) P(neither) = 0.125 (12.5%) +
+ +
+ Verification: +

Check: 0.375 (only Math) + 0.25 (both) + 0.25 (only Science) + 0.125 (neither) = 1.0 ✓

+
+
+ +
+

💪 Try These:

+
    +
  1. P(A)=0.6, P(B)=0.5, P(A∩B)=0.3. Find P(A∪B)
    2. +
  2. If P(A∪B)=0.8, P(A)=0.5, P(B)=0.4, find P(A∩B)
    3. +
  3. 100 students: 60 like pizza, 40 like burgers, 20 like both. How many like neither?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Two dice are rolled. Let A = "first die shows 6" and B = "sum is 7". Are A and B independent?

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Find P(A)

+
+ First die shows 6: one outcome out of 6
+ P(A) = 1/6 ≈ 0.167 +
+

Probability the first die is 6

+
+
+ +
+
Step 2:
+
+

Find P(B)

+
+ Sum equals 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
+ 6 favorable outcomes out of 36 total
+ P(B) = 6/36 = 1/6 ≈ 0.167 +
+

Count all ways to get sum of 7

+
+
+ +
+
Step 3:
+
+

Find P(A ∩ B)

+
+ First die is 6 AND sum is 7
+ Only possibility: (6,1)
+ P(A ∩ B) = 1/36 ≈ 0.028 +
+

Find where both events occur simultaneously

+
+
+ +
+
Step 4:
+
+

Test Independence

+
+ If independent: P(A ∩ B) = P(A) × P(B)
+ P(A) × P(B) = (1/6) × (1/6) = 1/36
+ P(A ∩ B) = 1/36
+ 1/36 = 1/36 ✓ EQUAL! +
+

Compare the two probabilities to test independence

+
+
+ +
+
Step 5:
+
+

Conclusion

+
+ Events A and B ARE independent
+ Knowing first die is 6 doesn't change probability of sum being 7 +
+

When the product rule holds, events are independent

+
+
+ +
+ ✓ Final Answer: + YES, events are independent. P(A∩B) = P(A)×P(B) = 1/36 +
+ +
+ Check: +

We can also verify: P(B|A) = P(A∩B)/P(A) = (1/36)/(1/6) = 1/6 = P(B). Since P(B|A) = P(B), the events are independent.

+
+
+ +
+

💪 Try These:

+
    +
  1. P(A)=0.3, P(B)=0.4, P(A∩B)=0.12. Independent?
    2. +
  2. Coin flip: P(Heads) and P(Tails). Independent?
    3. +
  3. Drawing two cards without replacement. Independent?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

A disease affects 1% of the population. A test is 99% accurate (detects 99% of sick people and correctly identifies 99% of healthy people). You test positive. What's the probability you actually have the disease?

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Define the Events and Given Information

+
+ Let A = has disease
+ Let B = tests positive
+ P(A) = 0.01 (1% of population has disease)
+ P(B|A) = 0.99 (99% true positive rate)
+ P(B|A') = 0.01 (1% false positive rate) +
+

Set up all known probabilities before applying Bayes' Theorem

+
+
+ +
+
Step 2:
+
+

Calculate P(B) using Total Probability

+
+ P(B) = P(B|A) × P(A) + P(B|A') × P(A')
+ P(B) = (0.99 × 0.01) + (0.01 × 0.99)
+ P(B) = 0.0099 + 0.0099 = 0.0198 +
+

Find the overall probability of testing positive

+
+
+ +
+
Step 3:
+
+

Apply Bayes' Theorem

+
+ P(A|B) = [P(B|A) × P(A)] / P(B)
+ P(A|B) = (0.99 × 0.01) / 0.0198
+ P(A|B) = 0.0099 / 0.0198
+ P(A|B) = 0.5 = 50% +
+

This is the posterior probability - what we want to find!

+
+
+ +
+ Final Answer: + Only 50% chance you have the disease despite testing positive! +
+ +
+ ✓ Why So Low? +

This counter-intuitive result occurs because the disease is so rare (1%). Even with a 99% accurate test, there are many more false positives from the healthy 99% than true positives from the sick 1%. Base rates matter!

+
+
+ +
+

💪 Try These:

+
    +
  1. What if the disease affects 10% of the population instead? Recalculate P(A|B)
    2. +
  2. If the test was 95% accurate instead of 99%, what would P(A|B) be?
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Continuous random variable X has uniform distribution on interval [0, 10]. a) Find the PDF f(x), b) Calculate P(3 ≤ X ≤ 7)

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Understand Uniform Distribution

+
+ X is equally likely anywhere between 0 and 10
+ For uniform on [a, b], PDF is constant
+ Total area under curve must equal 1 +
+

Uniform means constant probability density across the interval

+
+
+ +
+
Step 2:
+
+

Find PDF Height

+
+ Interval length = b - a = 10 - 0 = 10
+ For area = 1: height × width = 1
+ height × 10 = 1
+ height = 1/10 = 0.1
+ Therefore: f(x) = 0.1 for 0 ≤ x ≤ 10, and 0 otherwise +
+

The constant height must give total area of 1

+
+
+ +
+
Step 3:
+
+

Calculate P(3 ≤ X ≤ 7)

+
+ For continuous uniform: P(a ≤ X ≤ b) = (b-a) × height
+ P(3 ≤ X ≤ 7) = (7-3) × 0.1
+ = 4 × 0.1
+ = 0.4 +
+

Probability is the area of the rectangle

+
+
+ +
+
Step 4:
+
+

Visualize (Area Under Curve)

+
+ Rectangle: width = 4, height = 0.1
+ Area = 4 × 0.1 = 0.4
+ This represents probability +
+

The geometric area equals the probability

+
+
+ +
+ ✓ Final Answer: + a) f(x) = 0.1 for x ∈ [0,10]
b) P(3 ≤ X ≤ 7) = 0.4 (40%) +
+ +
+ Verification: +

P(0 ≤ X ≤ 10) = 10 × 0.1 = 1.0 ✓ (total probability = 1)

+
+
+ +
+

💪 Try These:

+
    +
  1. Uniform on [5,15]. Find PDF.
    2. +
  2. For above, find P(8 ≤ X ≤ 12)
    3. +
  3. Why is P(X = 7) = 0 for continuous distributions?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

For the uniform distribution from Topic 20 (X ~ Uniform[0,10]), find: a) F(5) = P(X ≤ 5), b) F(12), c) P(2 < X ≤ 8)

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Recall PDF

+
+ f(x) = 0.1 for 0 ≤ x ≤ 10
+ CDF is cumulative (area from left up to x) +
+

CDF accumulates probability from the left

+
+
+ +
+
Step 2:
+
+

Find F(5)

+
+ F(5) = P(X ≤ 5)
+ Area from 0 to 5: width = 5, height = 0.1
+ F(5) = 5 × 0.1 = 0.5 +
+

Half of the distribution is below x = 5

+
+
+ +
+
Step 3:
+
+

Find F(12)

+
+ F(12) = P(X ≤ 12)
+ But X can't exceed 10
+ All probability is accounted for by x = 10
+ F(12) = 1.0 (certainty) +
+

CDF plateaus at 1 beyond the support of the distribution

+
+
+ +
+
Step 4:
+
+

Find P(2 < X ≤ 8)

+
+ Using CDF: P(a < X ≤ b) = F(b) - F(a)
+ F(8) = 8 × 0.1 = 0.8
+ F(2) = 2 × 0.1 = 0.2
+ P(2 < X ≤ 8) = 0.8 - 0.2 = 0.6 +
+

Subtract lower CDF from upper CDF

+
+
+ +
+
Step 5:
+
+

General CDF Formula

+
+ For uniform [0, 10]:
+ • F(x) = 0 if x < 0
+ • F(x) = x/10 if 0 ≤ x ≤ 10
+ • F(x) = 1 if x > 10 +
+

The complete CDF function has three pieces

+
+
+ +
+ ✓ Final Answer: + a) F(5) = 0.5
b) F(12) = 1.0
c) P(2 < X ≤ 8) = 0.6 +
+ +
+ Check: +

F(0) = 0 (no probability below 0), F(10) = 1 (all probability by 10), F is non-decreasing ✓

+
+
+ +
+

💪 Try These:

+
    +
  1. For uniform [5,15], find F(10)
    2. +
  2. What is P(X > 7) using the CDF?
    3. +
  3. If F(x) = 0.75, what does this mean?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Flip a fair coin once. Let X = 1 if Heads, X = 0 if Tails. a) Find P(X=1) and P(X=0), b) Calculate E(X) and Var(X)

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Identify Bernoulli Trial

+
+ Single trial with two outcomes (Success/Failure)
+ Success = Heads, p = 0.5
+ Failure = Tails, 1-p = 0.5 +
+

This is a classic Bernoulli trial

+
+
+ +
+
Step 2:
+
+

Find Probabilities

+
+ P(X = 1) = p = 0.5 (probability of heads)
+ P(X = 0) = 1-p = 0.5 (probability of tails)
+ Check: 0.5 + 0.5 = 1.0 ✓ +
+

Probabilities must sum to 1

+
+
+ +
+
Step 3:
+
+

Calculate Expected Value

+
+ Formula: E(X) = p
+ E(X) = 0.5
+ Or: E(X) = 0×P(X=0) + 1×P(X=1)
+ = 0×0.5 + 1×0.5 = 0.5 ✓ +
+

Expected value is the probability of success

+
+
+ +
+
Step 4:
+
+

Calculate Variance

+
+ Formula: Var(X) = p(1-p)
+ Var(X) = 0.5 × 0.5 = 0.25
+ Standard deviation: σ = √0.25 = 0.5 +
+

Variance measures spread of outcomes

+
+
+ +
+
Step 5:
+
+

Interpret

+
+ On average, we get 0.5 heads per flip
+ Variance measures spread of 0 and 1 outcomes +
+

Expected value represents long-run average

+
+
+ +
+ ✓ Final Answer: + a) P(X=1) = 0.5, P(X=0) = 0.5
b) E(X) = 0.5, Var(X) = 0.25 +
+ +
+ Check: +

For fair coin, p = 0.5 makes sense. Over many flips, we expect half heads (E(X) = 0.5).

+
+
+ +
+

💪 Try These:

+
    +
  1. Biased coin: P(Heads) = 0.7. Find E(X) and Var(X)
    2. +
  2. Free throw: 80% success rate. Model as Bernoulli
    3. +
  3. When is Var(X) maximized for Bernoulli?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

IQ scores follow Normal distribution with μ = 100, σ = 15. Find: a) P(IQ ≤ 115), b) P(85 ≤ IQ ≤ 115), c) IQ score at 95th percentile

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Understand Normal Distribution

+
+ Bell-shaped, symmetric around mean
+ μ = 100 (center)
+ σ = 15 (spread) +
+

Parameters define the shape and location of the curve

+
+
+ +
+
Step 2:
+
+

Find P(IQ ≤ 115) using z-score

+
+ z = (x - μ)/σ = (115 - 100)/15 = 15/15 = 1
+ P(Z ≤ 1) = 0.8413 (from z-table)
+ About 84.13% have IQ ≤ 115 +
+

Standardize to z-score, then use standard normal table

+
+
+ +
+
Step 3:
+
+

Find P(85 ≤ IQ ≤ 115)

+
+ Lower bound: z₁ = (85-100)/15 = -15/15 = -1
+ Upper bound: z₂ = (115-100)/15 = 1
+ This is μ ± 1σ (68-95-99.7 rule)
+ P(-1 ≤ Z ≤ 1) = 0.68 (approximately 68%)
+ Exact: P(Z≤1) - P(Z≤-1) = 0.8413 - 0.1587 = 0.6826 +
+

One standard deviation on each side covers 68% of data

+
+
+ +
+
Step 4:
+
+

Find 95th Percentile

+
+ P(IQ ≤ x) = 0.95
+ From z-table: z = 1.645 for 95th percentile
+ x = μ + zσ = 100 + 1.645×15
+ = 100 + 24.675 = 124.675
+ IQ ≈ 125 +
+

Convert z-score back to original scale using inverse formula

+
+
+ +
+ ✓ Final Answer: + a) P(IQ ≤ 115) = 0.8413 (84.13%)
b) P(85 ≤ IQ ≤ 115) = 0.6826 (68.26%)
c) 95th percentile = IQ of 125 +
+ +
+ Verification: +

Using 68-95-99.7 rule: μ±1σ contains 68% ✓, μ±2σ contains 95%, μ±3σ contains 99.7%. Our answer matches the empirical rule!

+
+
+ +
+

💪 Try These:

+
    +
  1. Find P(IQ > 130) using same distribution
    2. +
  2. What IQ scores contain the middle 95% of people?
    3. +
  3. If z = -2, what percentile is this?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Explain the difference between α = 0.05 and α = 0.01. Which is more strict? Find critical values for both in a two-tailed test.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Understand α = 0.05

+
+ α = 0.05 means 5% significance
+ 95% confidence level (1 - 0.05)
+ P(Type I error) = 5%
+ Willing to be wrong 5% of the time +
+
+
+ +
+
Step 2:
+
+

Understand α = 0.01

+
+ α = 0.01 means 1% significance
+ 99% confidence level (1 - 0.01)
+ P(Type I error) = 1%
+ Only willing to be wrong 1% of the time +
+
+
+ +
+
Step 3:
+
+

Find Critical Values for α = 0.05

+
+ Two-tailed: split α into both tails
+ Each tail = 0.05/2 = 0.025
+ Z₀.₉₇₅ = ±1.96
+ Reject if |z| > 1.96 +
+
+
+ +
+
Step 4:
+
+

Find Critical Values for α = 0.01

+
+ Two-tailed: each tail = 0.01/2 = 0.005
+ Z₀.₉₉₅ = ±2.576
+ Reject if |z| > 2.576
+ Harder to reject (more strict!) +
+
+
+ +
+
Step 5:
+
+

Compare

+
+ α = 0.01 is MORE STRICT
+ Requires stronger evidence to reject H₀
+ Reduces Type I errors but increases Type II +
+
+
+ +
+ ✓ Final Answer: + α = 0.05: z = ±1.96; α = 0.01: z = ±2.576 (more strict) +
+
+ +
+

💪 Practice Problems:

+
    +
  1. Find critical value for α = 0.10, two-tailed
    2. +
  2. If we want to be very strict, should we use α = 0.05 or α = 0.001?
    3. +
  3. What happens to Type II error when α decreases?
    4. +
+
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Population has σ = 20. Calculate standard error for sample sizes: n = 4, n = 16, n = 64, n = 100. What pattern do you notice?

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Recall Standard Error Formula

+
+ SE = σ / √n
+ Where:
+ - σ = population standard deviation
+ - n = sample size
+ SE measures variability of sample means +
+
+
+ +
+
Step 2:
+
+

Calculate SE for n = 4

+
+ SE = 20 / √4
+ SE = 20 / 2
+ SE = 10 +
+
+
+ +
+
Step 3:
+
+

Calculate SE for n = 16

+
+ SE = 20 / √16
+ SE = 20 / 4
+ SE = 5 +
+
+
+ +
+
Step 4:
+
+

Calculate SE for n = 64

+
+ SE = 20 / √64
+ SE = 20 / 8
+ SE = 2.5 +
+
+
+ +
+
Step 5:
+
+

Calculate SE for n = 100

+
+ SE = 20 / √100
+ SE = 20 / 10
+ SE = 2 +
+
+
+ +
+
Step 6:
+
+

Analyze Pattern

+
+ n = 4: SE = 10
+ n = 16: SE = 5 (4× sample → ½ SE)
+ n = 64: SE = 2.5 (16× sample → ¼ SE)
+ n = 100: SE = 2 (25× sample → ⅕ SE)
+
+ Pattern: Quadruple sample size → Half the SE
+ Larger samples give more precise estimates! +
+
+
+ +
+ ✓ Final Answer: + SE: 10, 5, 2.5, 2. Larger n → Smaller SE (more precision) +
+
+ +
+

💪 Practice Problems:

+
    +
  1. If σ = 15 and n = 25, find SE
    2. +
  2. To cut SE in half, by what factor must we increase n?
    3. +
  3. Why does larger sample size reduce SE?
    4. +
+
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

A factory claims μ = 100. Sample: n = 36, x̄ = 105, σ = 12. Test at α = 0.05 (two-tailed).

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

State Hypotheses

+
+ H₀: μ = 100 (claim is true)
+ H₁: μ ≠ 100 (claim is false)
+ α = 0.05, two-tailed test +
+
+
+ +
+
Step 2:
+
+

Calculate Standard Error

+
+ SE = σ / √n
+ SE = 12 / √36
+ SE = 12 / 6
+ SE = 2 +
+
+
+ +
+
Step 3:
+
+

Calculate Z-Statistic

+
+ z = (x̄ - μ₀) / SE
+ z = (105 - 100) / 2
+ z = 5 / 2
+ z = 2.5 +
+
+
+ +
+
Step 4:
+
+

Find Critical Values

+
+ α = 0.05, two-tailed
+ Critical values: z = ±1.96
+ Rejection regions: z < -1.96 or z > 1.96 +
+
+
+ +
+
Step 5:
+
+

Make Decision

+
+ Test statistic: z = 2.5
+ Critical value: z = 1.96
+ 2.5 > 1.96 → In rejection region
+
+ REJECT H₀ +
+
+
+ +
+
Step 6:
+
+

Interpret

+
+ There IS significant evidence that μ ≠ 100
+ The sample mean of 105 is statistically different
+ Factory's claim is likely false +
+
+
+ +
+ ✓ Final Answer: + z = 2.5 > 1.96, REJECT H₀ (claim is false) +
+ +
+ Check: +

P-value = 2 × P(Z > 2.5) = 2 × 0.0062 = 0.0124 < 0.05 ✓ Confirms rejection

+
+
+ +
+

💪 Practice Problems:

+
    +
  1. Test: μ₀ = 50, x̄ = 48, σ = 10, n = 25, α = 0.05
    2. +
  2. If z = -1.5, α = 0.05, two-tailed, what's your decision?
    3. +
  3. When should we use z-test vs t-test?
    4. +
+
+
+

🎯 Key Takeaways

- - -
-

Critical Values

- +
+ +
+

Critical Values

+ +
+ + +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Find critical z-values for: a) α = 0.05 one-tailed (right), b) α = 0.05 two-tailed, c) α = 0.01 two-tailed. Draw rejection regions.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

One-Tailed Right (α = 0.05)

+
+ All α in right tail
+ Find z where P(Z > z) = 0.05
+ P(Z ≤ z) = 1 - 0.05 = 0.95
+ From z-table: z₀.₉₅ = 1.645
+
+ Critical value: z = 1.645
+ Reject H₀ if z > 1.645 +
+
+
+ +
+
Step 2:
+
+

Two-Tailed (α = 0.05)

+
+ Split α between both tails
+ Each tail = 0.05/2 = 0.025
+ Left tail: P(Z < z) = 0.025 → z = -1.96
+ Right tail: P(Z > z) = 0.025 → z = +1.96
+
+ Critical values: z = ±1.96
+ Reject H₀ if |z| > 1.96 +
+
+
+ +
+
Step 3:
+
+

Two-Tailed (α = 0.01)

+
+ More strict test
+ Each tail = 0.01/2 = 0.005
+ P(Z < z) = 0.005 → z = -2.576
+ P(Z > z) = 0.005 → z = +2.576
+
+ Critical values: z = ±2.576
+ Reject H₀ if |z| > 2.576 +
+
+
+ +
+
Step 4:
+
+

Visualize Rejection Regions

+
+ One-tailed (α=0.05): [______|████] z > 1.645
+ Two-tailed (α=0.05): [██|________|██] |z| > 1.96
+ Two-tailed (α=0.01): [█|__________|█] |z| > 2.576
+
+ Smaller α → Larger critical values → Harder to reject +
+
+
+ +
+ ✓ Final Answer: + a) z = 1.645, b) z = ±1.96, c) z = ±2.576 +
+
+ +
+

💪 Practice Problems:

+
    +
  1. Find critical value for α = 0.10, one-tailed (left)
    2. +
  2. If your test statistic is z = 2.0, which tests would reject H₀?
    3. +
  3. Why are two-tailed critical values larger than one-tailed?
    4. +
+
@@ -1914,6 +3662,113 @@

P-value is NOT the probability that H₀ is true! It's the probability of observing your data IF H₀ were true.

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Sample of 36 students has mean score x̄ = 78. Population mean claimed to be μ₀ = 75 with σ = 12. Test at α = 0.05 using p-value method.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

State Hypotheses

+
+ H₀: μ = 75 (null hypothesis - no difference)
+ H₁: μ ≠ 75 (alternative - there is a difference)
+ Two-tailed test +
+

Set up null and alternative hypotheses

+
+
+ +
+
Step 2:
+
+

Calculate Test Statistic

+
+ z = (x̄ - μ₀) / (σ/√n)
+ z = (78 - 75) / (12/√36)
+ z = 3 / (12/6)
+ z = 3 / 2 = 1.5 +
+

Calculate the z-score

+
+
+ +
+
Step 3:
+
+

Find P-Value

+
+ For two-tailed: p-value = 2 × P(Z > |1.5|)
+ P(Z > 1.5) = 1 - 0.9332 = 0.0668
+ p-value = 2 × 0.0668 = 0.1336 +
+

Multiply by 2 for two-tailed test

+
+
+ +
+
Step 4:
+
+

Compare with α

+
+ p-value = 0.1336
+ α = 0.05
+ 0.1336 > 0.05 +
+

Since p-value exceeds α, we fail to reject H₀

+
+
+ +
+
Step 5:
+
+

Make Decision

+
+ Since p-value > α, FAIL TO REJECT H₀
+ Not enough evidence to conclude mean differs from 75
+ p-value of 13.36% means we'd see results this extreme
+ 13.36% of time if H₀ true +
+

Interpret in context

+
+
+ +
+ ✓ Final Answer: + p-value = 0.1336 > 0.05, Fail to reject H₀ +
+ +
+ Check: +

The result is not statistically significant at α = 0.05 level. We need stronger evidence to claim the mean differs from 75.

+
+
+ +
+

💪 Try These:

+
    +
  1. If z = 2.5, α = 0.01, find p-value and decide
    2. +
  2. When do we reject H₀ using p-value method?
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Researcher claims new drug LOWERS blood pressure (μ < 120). Sample of 49: x̄ = 115, σ = 21. Test at α = 0.05. Should this be one-tailed or two-tailed?

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Analyze the Claim

+
+ Claim: drug LOWERS pressure (directional)
+ Looking for decrease specifically
+ This requires ONE-TAILED test (left tail) +
+

Directional claim = one-tailed test

+
+
+ +
+
Step 2:
+
+

Set Up Hypotheses

+
+ H₀: μ ≥ 120 (blood pressure not lower)
+ H₁: μ < 120 (blood pressure IS lower)
+ Left-tailed test +
+

Alternative hypothesis shows the direction

+
+
+ +
+
Step 3:
+
+

Calculate Z-Score

+
+ z = (x̄ - μ₀) / (σ/√n)
+ z = (115 - 120) / (21/√49)
+ z = -5 / (21/7)
+ z = -5 / 3 = -1.67 +
+

Negative z-score indicates below mean

+
+
+ +
+
Step 4:
+
+

Find Critical Value (One-Tailed)

+
+ For α = 0.05, one-tailed (left)
+ Critical value: z = -1.645 +
+

One-tailed critical value differs from two-tailed

+
+
+ +
+
Step 5:
+
+

Make Decision

+
+ Test statistic: z = -1.67
+ Critical value: z = -1.645
+ -1.67 < -1.645 (in rejection region)
+ REJECT H₀ +
+

Falls in rejection region, so reject null

+
+
+ +
+
Step 6:
+
+

Contrast with Two-Tailed

+
+ If two-tailed: critical values ±1.96
+ Our |z| = 1.67 < 1.96
+ Would NOT reject H₀ with two-tailed!
+ This shows importance of choosing correct test +
+

Test choice matters!

+
+
+ +
+ ✓ Final Answer: + Use ONE-TAILED (left). z = -1.67 < -1.645, Reject H₀ +
+ +
+ Check: +

Evidence supports claim that drug lowers blood pressure. One-tailed test was appropriate for directional claim.

+
+
+ +
+

💪 Try These:

+
    +
  1. Claim: μ > 50. One-tailed or two-tailed?
    2. +
  2. Claim: μ ≠ 100. Which test?
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Small sample: n = 16, x̄ = 52, s = 8. Test if μ = 50 at α = 0.05. Population σ unknown.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Choose Correct Test

+
+ n = 16 < 30 (small sample)
+ σ unknown (use sample s)
+ Use T-TEST instead of z-test +
+

Small sample + unknown σ = t-test

+
+
+ +
+
Step 2:
+
+

Calculate T-Statistic

+
+ t = (x̄ - μ₀) / (s/√n)
+ t = (52 - 50) / (8/√16)
+ t = 2 / (8/4)
+ t = 2 / 2 = 1.0 +
+

Use sample standard deviation s

+
+
+ +
+
Step 3:
+
+

Find Degrees of Freedom

+
+ df = n - 1
+ df = 16 - 1 = 15 +
+

Lose 1 df for estimating mean

+
+
+ +
+
Step 4:
+
+

Find Critical Value

+
+ Two-tailed test, α = 0.05
+ df = 15
+ From t-table: t₀.₀₂₅,₁₅ = ±2.131 +
+

Look up in t-distribution table

+
+
+ +
+
Step 5:
+
+

Compare and Decide

+
+ Test statistic: t = 1.0
+ Critical values: ±2.131
+ |1.0| < 2.131
+ FAIL TO REJECT H₀ +
+

Test statistic not in rejection region

+
+
+ +
+
Step 6:
+
+

Interpret

+
+ Not enough evidence that μ ≠ 50
+ Sample mean of 52 is not significantly different from 50 +
+

Interpret in context of problem

+
+
+ +
+ ✓ Final Answer: + t = 1.0, critical = ±2.131, Fail to reject H₀ +
+ +
+ Check: +

The difference between 52 and 50 is not statistically significant at α = 0.05 level with this small sample.

+
+
+ +
+

💪 Try These:

+
    +
  1. n = 25, x̄ = 100, s = 15, test μ = 95 at α = 0.01
    2. +
  2. Why use t-test instead of z-test?
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Calculate degrees of freedom for: a) Single sample t-test: n = 20, b) Two-sample t-test: n₁ = 15, n₂ = 18, c) Chi-squared test: 3×4 contingency table

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Single Sample T-Test

+
+ Formula: df = n - 1
+ n = 20
+ df = 20 - 1 = 19
+ We "lose" 1 df because we estimate mean from sample +
+

Each parameter estimated reduces df by 1

+
+
+ +
+
Step 2:
+
+

Two-Sample T-Test (Equal Variances)

+
+ Formula: df = n₁ + n₂ - 2
+ n₁ = 15, n₂ = 18
+ df = 15 + 18 - 2 = 31
+ Lose 1 df per sample for estimating each mean +
+

Two samples = two means estimated

+
+
+ +
+
Step 3:
+
+

Chi-Squared Contingency Table

+
+ Formula: df = (rows - 1) × (columns - 1)
+ 3 rows, 4 columns
+ df = (3 - 1) × (4 - 1)
+ df = 2 × 3 = 6 +
+

Degrees of freedom for independence test

+
+
+ +
+
Step 4:
+
+

Explain Concept

+
+ Degrees of freedom = number of values free to vary
+ Each parameter estimated reduces df by 1
+ Higher df → distribution closer to normal +
+

Conceptual understanding

+
+
+ +
+ ✓ Final Answer: + a) df = 19, b) df = 31, c) df = 6 +
+ +
+ Check: +

These df values would be used to find appropriate critical values from respective distribution tables.

+
+
+ +
+

💪 Try These:

+
    +
  1. Sample size 100, find df for t-test
    2. +
  2. 5×3 table, find df for chi-squared
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Drug trial tests H₀: "Drug is safe" vs H₁: "Drug is dangerous". Describe Type I and Type II errors with consequences.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Define Type I Error (False Positive)

+
+ Type I: Reject H₀ when H₀ is TRUE
+ In this case: Conclude drug is dangerous when it's actually safe
+ Probability = α (significance level)
+ Consequence: Safe drug rejected, patients miss beneficial treatment +
+

False alarm - reject truth

+
+
+ +
+
Step 2:
+
+

Define Type II Error (False Negative)

+
+ Type II: Fail to reject H₀ when H₁ is TRUE
+ In this case: Conclude drug is safe when it's actually dangerous
+ Probability = β
+ Consequence: Dangerous drug approved, patients harmed! +
+

Miss detecting danger

+
+
+ +
+
Step 3:
+
+

Create Decision Matrix

+
+ Reality vs Decision:
+ If H₀ true (safe) + Reject H₀ (call dangerous) = TYPE I
+ If H₁ true (dangerous) + Fail to reject = TYPE II
+ Correct decisions: Accept truth or reject false +
+

Four possible outcomes

+
+
+ +
+
Step 4:
+
+

Calculate Example

+
+ If α = 0.05: 5% chance of Type I error
+ If β = 0.20: 20% chance of Type II error
+ Power = 1 - β = 0.80 (80% chance of detecting dangerous drug) +
+

Probabilities of each error

+
+
+ +
+
Step 5:
+
+

Compare Consequences

+
+ Type I: Waste safe drug (economic cost)
+ Type II: Approve dangerous drug (LIFE RISK!)
+ Type II often more serious → use lower α +
+

Context determines which error is worse

+
+
+ +
+ ✓ Final Answer: + Type I (α): Reject safe drug
Type II (β): Approve dangerous drug
Type II more dangerous in this case! +
+ +
+ Check: +

In medical contexts, Type II errors (missing danger) are often considered worse than Type I errors (false alarms).

+
+
+ +
+

💪 Try These:

+
    +
  1. Security scanner: H₀ = "Safe". Describe Type I/II errors
    2. +
  2. If α = 0.01, what's P(Type I error)?
    3. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Find the derivative of f(x) = 3x⁴ - 2x³ + 5x - 7

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Identify the Power Rule

+
+ Power Rule: d/dx(xⁿ) = n·xⁿ⁻¹
+ Apply to each term separately +
+

The power rule works term by term

+
+
+ +
+
Step 2:
+
+

Derivative of First Term: 3x⁴

+
+ Coefficient stays: 3
+ Exponent: 4 → multiply by 4 and decrease exponent
+ Result: 3 × 4 × x³ = 12x³ +
+

Bring down the 4, multiply by coefficient 3

+
+
+ +
+
Step 3:
+
+

Derivative of Second Term: -2x³

+
+ Result: -2 × 3 × x² = -6x² +
+

Same process, keep the negative sign

+
+
+ +
+
Step 4:
+
+

Derivative of Third Term: 5x

+
+ 5x = 5x¹
+ Result: 5 × 1 × x⁰ = 5 +
+

x⁰ = 1, so we just get the coefficient

+
+
+ +
+
Step 5:
+
+

Derivative of Fourth Term: -7

+
+ Constant → derivative is 0 +
+

Constants disappear when we take the derivative

+
+
+ +
+
Step 6:
+
+

Combine All Terms

+
+ f'(x) = 12x³ - 6x² + 5 + 0
+ f'(x) = 12x³ - 6x² + 5 +
+

Sum the derivatives of each term

+
+
+ +
+ Final Answer: + f'(x) = 12x³ - 6x² + 5 +
+ +
+ ✓ Check: +

At x=1: f'(1) = 12(1) - 6(1) + 5 = 11. This is the slope of the tangent line at x=1.

+
+
+ +
+

💪 Try These:

+
    +
  1. Find f'(x) for f(x) = 2x³ + 4x² - 3
    2. +
  2. Find the derivative of g(x) = 5x⁴ - x
    3. +
  3. What is d/dx(7x² + 2x + 8)?
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Calculate the definite integral: ∫₀³ (2x + 1) dx

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Find the Antiderivative

+
+ ∫(2x + 1) dx
+ ∫2x dx = 2 × (x²/2) = x²
+ ∫1 dx = x
+ F(x) = x² + x + C +
+

Use the power rule in reverse

+
+
+ +
+
Step 2:
+
+

Apply the Fundamental Theorem of Calculus (Part 2)

+
+ ∫ₐᵇ f(x)dx = F(b) - F(a)
+ No need for +C (it cancels out) +
+

Evaluate antiderivative at bounds and subtract

+
+
+ +
+
Step 3:
+
+

Evaluate at Upper Bound (x = 3)

+
+ F(3) = 3² + 3
+ F(3) = 9 + 3 = 12 +
+

Substitute x = 3 into F(x)

+
+
+ +
+
Step 4:
+
+

Evaluate at Lower Bound (x = 0)

+
+ F(0) = 0² + 0
+ F(0) = 0 +
+

Substitute x = 0 into F(x)

+
+
+ +
+
Step 5:
+
+

Subtract: F(upper) - F(lower)

+
+ ∫₀³ (2x + 1) dx = F(3) - F(0)
+ = 12 - 0 = 12 +
+

Final calculation

+
+
+ +
+ Final Answer: + 12 +
+ +
+ ✓ Meaning: +

The area under the curve y = 2x + 1 from x = 0 to x = 3 is 12 square units. This represents the accumulated value of the function over that interval.

+
+
+ +
+

💪 Try These:

+
    +
  1. Calculate ∫₁⁴ x² dx
    2. +
  2. Find ∫₀² (3x² + 2) dx
    3. +
  3. Evaluate ∫₁³ (4x - 1) dx
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Find the best-fit line for the data points: (1,2), (2,4), (3,5), (4,7), (5,8)

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Calculate the Means

+
+ x̄ = (1+2+3+4+5)/5 = 15/5 = 3
+ ȳ = (2+4+5+7+8)/5 = 26/5 = 5.2 +
+

Find the average of x values and y values

+
+
+ +
+
Step 2:
+
+

Set Up Slope Formula

+
+ b = Σ(x-x̄)(y-ȳ) / Σ(x-x̄)²
+ Need to calculate deviations and their products +
+

This is the least squares formula for slope

+
+
+ +
+
Step 3:
+
+

Create Calculation Table

+
+ x | y | (x-x̄) | (y-ȳ) | (x-x̄)(y-ȳ) | (x-x̄)²
+ 1 | 2 | -2 | -3.2 | 6.4 | 4
+ 2 | 4 | -1 | -1.2 | 1.2 | 1
+ 3 | 5 | 0 | -0.2 | 0 | 0
+ 4 | 7 | 1 | 1.8 | 1.8 | 1
+ 5 | 8 | 2 | 2.8 | 5.6 | 4
+ Sum| | | | 15 | 10 +
+

Organize all calculations in a table

+
+
+ +
+
Step 4:
+
+

Calculate Slope (b)

+
+ b = 15 / 10 = 1.5 +
+

For every 1 unit increase in x, y increases by 1.5

+
+
+ +
+
Step 5:
+
+

Calculate Intercept (a)

+
+ a = ȳ - b·x̄
+ a = 5.2 - (1.5 × 3)
+ a = 5.2 - 4.5 = 0.7 +
+

Where the line crosses the y-axis

+
+
+ +
+
Step 6:
+
+

Write the Equation

+
+ y = a + bx
+ y = 0.7 + 1.5x +
+

This is our best-fit line!

+
+
+ +
+ Final Answer: + y = 0.7 + 1.5x +
+ +
+ ✓ Check: +

When x=3: y = 0.7 + 1.5(3) = 5.2 ✓ (matches our mean!)

+

When x=1: y = 0.7 + 1.5(1) = 2.2 (close to actual 2)

+
+
+ +
+

💪 Try These:

+
    +
  1. Fit a line to: (0,1), (1,3), (2,5)
    2. +
  2. Using y=2+3x, predict y when x=6
    3. +
  3. If b=2 and the line passes through (3,10), find a
    4. +
+ + +
+
+

🎯 Key Takeaways

+ +
+

📝 Worked Example - Step by Step

+ +
+

Problem:

+

Minimize f(x) = x² using gradient descent. Start at x₀ = 5, learning rate α = 0.1, run 3 iterations.

+
+ +
+

Solution:

+ +
+
Step 1:
+
+

Find the Gradient (Derivative)

+
+ f(x) = x²
+ f'(x) = 2x
+ This is the gradient we'll use +
+

The gradient tells us which direction is uphill

+
+
+ +
+
Step 2:
+
+

Iteration 1

+
+ Current position: x₀ = 5
+ Gradient: f'(5) = 2(5) = 10
+ Update: x₁ = x₀ - α·f'(x₀)
+ x₁ = 5 - 0.1(10) = 5 - 1 = 4 +
+

Move against the gradient (downhill)

+
+
+ +
+
Step 3:
+
+

Iteration 2

+
+ Current position: x₁ = 4
+ Gradient: f'(4) = 2(4) = 8
+ Update: x₂ = 4 - 0.1(8) = 4 - 0.8 = 3.2 +
+

Gradient is smaller, we're getting closer to minimum

+
+
+ +
+
Step 4:
+
+

Iteration 3

+
+ Current position: x₂ = 3.2
+ Gradient: f'(3.2) = 2(3.2) = 6.4
+ Update: x₃ = 3.2 - 0.1(6.4) = 3.2 - 0.64 = 2.56 +
+

Still moving toward x = 0 (the true minimum)

+
+
+ +
+
Step 5:
+
+

Summary of Convergence

+
+ Iter | x | f(x) | Gradient
+ 0 | 5.00 | 25.00 | 10.0
+ 1 | 4.00 | 16.00 | 8.0
+ 2 | 3.20 | 10.24 | 6.4
+ 3 | 2.56 | 6.55 | 5.12
+ ... | ... | ... | ...
+ ∞ | 0.00 | 0.00 | 0.0 +
+

If we continue, x converges to 0

+
+
+ +
+ Final Answer: + After 3 iterations: x₃ = 2.56, f(x₃) = 6.55
+ At convergence: x = 0, minimum value f(0) = 0 +
+ +
+ ✓ Check: +

The function f(x) = x² has its minimum at x = 0. Gradient descent successfully moves us from x = 5 toward x = 0. With more iterations (or larger learning rate), we'd get even closer!

+
+
+ +
+

💪 Try These:

+
    +
  1. Use GD to minimize f(x) = x² - 4x starting at x=0, α=0.1, 2 iterations
    2. +
  2. Find the minimum of f(x) = (x-3)² using calculus (no GD)
    3. +
  3. If gradient is positive, which direction should we move?
    4. +
+ + +
+
+

🎯 Key Takeaways