problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
334. A machine manufactures balls. A ball is considered suitable if the deviation $X$ of the ball's diameter from the design size in absolute value is less than 0.7 mm. Assuming that the random variable $X$ is normally distributed with a standard deviation $\sigma=0.4$ mm, find how many suitable balls on average will b... | Solution. Since $X$ is the deviation (of the diameter of the ball from the design size), then $M(X)=a=0$.
Let's use the formula $P(|X|<\delta)=2 \Phi(\delta / \sigma)$. Substituting $\delta=0.7, \sigma=0.4$, we get
$$
P(|X|<0.7)=2 \Phi\left(\frac{0.7}{0.4}\right)=2 \Phi(1.75)=2 \cdot 0.4599=0.92
$$
Thus, the probabi... | 92 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,396 |
337. A random variable $X$ is normally distributed with a mean of $a=10$. The probability of $X$ falling in the interval $(10,20)$ is 0.3. What is the probability of $X$ falling in the interval $(0,10)$? | Solution. Since the normal curve is symmetric about the line $x=a=10$, the areas bounded above by the normal curve and below by the intervals (0,10) and (10,20) are equal to each other. Since these areas are numerically equal to the probabilities of $X$ falling into the corresponding interval, then
$$
P(0<X<10)=P(10<X... | 0.3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,397 |
349. Prove that if a continuous random variable $X$ is distributed according to the exponential law, then the probability of $X$ falling into the interval ( $a, b$ ) is $\mathrm{e}^{-\lambda a}-\mathrm{e}^{-\lambda b}$ | Solution. First method. Let the magnitude $X$ be given by the distribution function $F(x)=1-\mathrm{e}^{-\lambda x}(x \geqslant 0)$. Then the probability of $X$ falling into the interval $(a, b)$ (see Chapter VI, § 1)
$$
P(a<X<b)=F(b)-F(a)=\left[1-\mathrm{e}^{-\lambda b}\right]-\left[1-\mathrm{e}^{-\lambda a}\right]=\... | \mathrm{e}^{-\lambda}-\mathrm{e}^{-\lambdab} | Calculus | proof | Yes | Yes | olympiads | false | 32,399 |
350. A continuous random variable $X$ is distributed according to the exponential law, given by the probability density function $f(x)=3 \mathrm{e}^{-3 x}$ for $x \geqslant 0$; for $x<0$, $f(x)=0$. Find the probability that in the experiment, $X$ falls into the interval $(0.13, 0.7)$. | Solution. We use the formula
$$
P(a<X<b)=\mathrm{e}^{-\lambda a}-\mathrm{e}^{-\lambda b} .
$$
Considering that, according to the condition, $a=0.13, b=0.7, \lambda=3$, and using the table of values of the function $\mathrm{e}^{-x}$, we get
$$
\begin{gathered}
P(0.13<X<0.7)=\mathrm{e}^{-3 \cdot 0.13}-\mathrm{e}^{-3 \... | 0.555 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,400 |
353. Find the mathematical expectation of the exponential distribution
$$
f(x)=\lambda \mathrm{e}^{-\lambda x} \quad(x \geqslant 0) ; \quad f(x)=0 \quad(x<0) .
$$ | Solution. We use the formula
$$
M(X)=\int_{-\infty}^{\infty} x f(x) \mathrm{d} x
$$
Considering that $f(x)=0$ for $x<0$ and $f(x)=\lambda \mathrm{e}^{-\lambda x}$ for $x \geqslant 0$, we get
$$
M(X)=\lambda \int_{0}^{\infty} x \cdot \mathrm{e}^{-\lambda x} \mathrm{~d} x
$$
Integrating by parts using the formula
$$... | \frac{1}{\lambda} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,401 |
356. Find: a) variance; b) standard deviation of the exponential distribution given by the probability density function: $f(x)=\lambda \mathrm{e}^{-\lambda x}$ for $x \geqslant 0 ; f(x)=0$ for $x<0$. | Solution. a) Using the formula
$$
D(X)=\int_{-\infty}^{\infty} x^{2} f(x) \mathrm{d} x-[M(X)]^{2}
$$
Considering that $f(x)=0$ for $x<0, M(X)=1 / \lambda$ (see problem 358), we get
$$
D(X)=\lambda \int_{0}^{\infty} x^{2} e^{-\lambda x} \mathrm{~d} x-(1 / \lambda)^{2}
$$
Integrating by parts twice, we find
$$
\lamb... | D(X)=\frac{1}{\lambda^2},\sigma(X)=\frac{1}{\lambda} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,402 |
364. Prove that a continuous random variable $T$ - the time between the occurrences of two consecutive events in a simple stream with a given intensity $\lambda$ (see Chapter IV, § 2) - has an exponential distribution $F(t)=1-e^{-\lambda t}(t \geqslant 0)$. | Solution. Suppose that at time $t_{0}$, event $A_{i}$ of the flow occurred. Let $t_{1}=t_{0}+t$ (it is recommended to draw a time axis and mark points $t_{0}$ and $t_{1}$ for clarity).
If at least one event of the flow following event $\boldsymbol{A}_{\mathbf{i}}$ occurs within the interval enclosed in the interval ( ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,403 |
367. The duration of trouble-free operation of an element has an exponential distribution $F(t)=1$ - $\mathrm{e}^{-0.01 t}(t>0)$. Find the probability that during a time period of $t=50$ hours: a) the element will fail; b) the element will not fail. | Solution. a) Since the distribution function $F(t)=1-\mathrm{e}^{-0.01 t}$ defines the probability of the element failing within a time duration $t$, substituting $t=50$ into the distribution function, we get the probability of failure:
$$
F(50)=1-\mathrm{e}^{-0.01 \cdot 50}=1-\mathrm{e}^{-0.3}=1-0.606=0.394 ;
$$
b) ... | 0.3940.606 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,404 |
372. The exponential reliability law refers to the reliability function defined by the equation $R(t)=\mathrm{e}^{-\lambda t}$, where the positive number $\lambda$ is the failure intensity. Prove the characteristic property of the exponential reliability law: the probability of fault-free operation of an element over a... | Solved. Let's introduce the notations of events: $A$ - faultless operation of the element in the interval $(0, t_{0})$ of duration $t_{0}$; $B$ - faultless operation of the element in the interval $(t_{0}, t_{0}+t)$ of duration $t$.
Then $AB$ - faultless operation in the interval $(0, t_{0}+t)$ of duration $t_{0}+t$.
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,405 |
373. A discrete random variable $X$ is given by the distribution law:
$$
\begin{array}{cccc}
X & 1 & 3 & 5 \\
p & 0.4 & 0.1 & 0.5
\end{array}
$$
Find the distribution law of the random variable $Y=3X$. | Solution. Let's find the possible values of the quantity $Y=3X$. We have: $y_{1}=3\cdot1=3; y_{2}=3\cdot3=9; y_{n}=3\cdot5=15$. We see that different possible values of $X$ correspond to different values of $Y$. This is explained by the fact that the function $y=\varphi(x)=3x$ is monotonic. Let's find the probabilities... | \begin{pmatrix}Y&3&9&15\\p&0.4&0.1&0.5\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,406 |
375. A discrete random variable $X$ is given by the distribution law:
$$
\begin{array}{lrrrr}
X & -1 & -2 & 1 & 2 \\
p & 0.3 & 0.1 & 0.2 & 0.4
\end{array}
$$
Find the distribution law of the random variable $Y=X^{2}$. | Solution. Let's find the possible values of $Y$:
$$
\begin{array}{cc}
y_{i}=x_{1}^{2}=(-1)^{2}=1, & y_{2}=x_{2}^{2}=(-2)^{2}=4, \\
y_{3}=x_{3}^{2}=1^{2}=1, & y_{4}=x_{4}^{2}=2^{2}=4 .
\end{array}
$$
Thus, different values of $X$ correspond to the same values of $Y$. This is explained by the fact that the possible val... | \begin{pmatrix}Y&1&4\\p&0.5&0.5\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,407 |
377. The density function $f(x)$ of a random variable $X$, whose possible values are contained in the interval $(a, b)$, is given. Find the density function of the random variable $Y=3X$. | S o l u t i o n. Since the function $y=3 x$ is differentiable and strictly increasing, the formula
$$
g(y)=f[\psi(y)] \cdot\left|\psi^{\prime}(y)\right|
$$
can be applied, where $\Psi(y)$ is the inverse function of $y=3 x$.
Let's find $\psi(y):$
Let's find $f[\psi(y)]:$
$$
\psi(y)=x=y / 3
$$
$$
f[\psi(y)]=f(y / 3... | (y)=\frac{1}{3}f(y/3) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,408 |
382. In a rectangular coordinate system $x O y$, a ray is randomly drawn (at an arbitrary angle $t$) from the point $A(4 ; 0)$, intersecting the $O y$ axis. Find the density $g(y)$ of the probability distribution of the ordinate $y$ of the intersection point of the drawn ray with the $O y$ axis. | Solution. The angle $t$ can be considered as a random variable uniformly distributed in the interval ($-\pi / 2, \pi / 2$), and in this interval, the probability density function is
$$
f(t)=\frac{1}{\pi / 2-(-\pi / 2)}=\frac{1}{\pi}:
$$
outside the considered interval, $f(t)=0$.
From Fig. 7, it follows that the ordi... | (y)=\frac{4}{\pi(16+y^{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,409 |
383. A random variable $X$ is uniformly distributed in the interval $(-\pi / 2, \pi / 2)$. Find the density function $g(y)$ of the random variable $Y=\sin X$. | Solution. Let's find the density function $f(x)$ of the random variable $X$. The variable $X$ is uniformly distributed in the interval $(-\pi / 2, \pi / 2)$, so in this interval
$$
f(x)=\frac{1}{\pi / 2-(-\pi / 2)}=\frac{1}{\pi}
$$
outside the considered interval, $f(x)=0$.
The function $y=\sin x$ is monotonic in th... | (y)=\frac{1}{\pi\sqrt{1-y^{2}}},\quad-1<y<1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,410 |
386. A random variable $X$ is uniformly distributed in the interval $(0,2 \pi)$. Find the density function $g(y)$ of the random variable $Y=\cos X$. | Solution. Let's find the density function $f(x)$ of the random variable $X$: in the interval $(0, 2\pi)$ we have
$$
f(x)=1/(2\pi-0)=1/2\pi
$$
outside this interval, $f(x)=0$.
From the equation $y=\cos x$, let's find the inverse function $x=\psi(y)$. Since the function $y=\cos x$ is not monotonic in the interval $(0,... | (y)=\frac{1}{\pi\sqrt{1-y^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,411 |
388. The random variable $X$ is normally distributed with a mathematical expectation equal to $a$, and a standard deviation equal to $\sigma$. Prove that the linear function $Y=A X+B$ is also normally distributed, and
$$
M(Y)=A a+B, \quad \sigma(Y)=|A| \sigma
$$ | Solution. Let's write the density function of the random variable $X$:
$$
f(x)=\frac{1}{\sigma \sqrt{2 \pi}} \mathrm{e}^{-(x-a)^{2} / 2 \sigma^{2}}
$$
The function $y=A x+B$ is monotonic, so we can use the formula
$$
g(y)=f[\psi(y)] \cdot\left|\psi^{\prime}(y)\right| .
$$
Let's find $x=\psi(y)$ from the equation $y... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,412 |
389. Given the density function $f(x)=\frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-x^{2} / 2},(-\infty<x<\infty)$ of a normally distributed random variable $X$. Find the density function $g(y)$ of the random variable $Y=X^{2}$. | Solution. From the equation $y=x^{2}$, we find the inverse function. Since in the interval $(-\infty, \infty)$ the function $y=x^{2}$ is not monotonic, we divide this interval into intervals $(-\infty, 0)$ and $(0, \infty)$, in which the considered function is monotonic. In the interval $(-\infty, 0)$, the inverse func... | (y)=\frac{1}{\sqrt{2\piy}}e^{-y/2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,413 |
392. A random variable $X$ is given by the probability density function $f(x)=(1 / 2) \sin x$ in the interval $(0, \pi)$; outside this interval, $f(x)=0$. Find the mathematical expectation of the random variable $Y=\varphi(X)=X^{2}$, by first determining the density function $g(Y)$ of the variable $Y$.
| Solution. First, we find the density $g(y)$ of the random variable $Y$. Since the function $y=\varphi(x)=x^{2}$ is strictly increasing for the considered values of $x (0<x<\pi)$, the density $g(y)$ will be found using the formula
$$
g(y)=f[\psi(y)] \cdot\left|\psi^{\prime}(y)\right|
$$
where $\Psi(y)=\sqrt{y}$ is the... | (\pi^{2}-4) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,414 |
394. The random variable $X$ is given by the probability density function $f(x)=(1 / 2) \sin x$ in the interval $(0, \pi)$; outside this interval, $f(x)=0$. Find the variance of the function $Y=\varphi(X)=X^{2}$, using the density function $g(y)$. | Solution. We use the formula
$$
D(Y)=\int_{c}^{d} y^{2} g(y) \mathrm{d} y-[M(Y)]^{2}
$$
where $\boldsymbol{c}$ and $d$ are the endpoints of the interval in which the possible values of $Y$ are contained. Substituting $g(y)=\sin \sqrt{y} / 4 \sqrt{y}, M(Y)=\left(\pi^{2}-4\right) / 2$ (see problem 392) and noting that ... | (\pi^{4}-16\pi^{2}+80)/4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,415 |
397. Given the distribution function $F(x)$ of a random variable $X$. Find the distribution function $G(y)$ of the random variable $Y=3X+2$.
翻译结果如下：
397. Given the distribution function $F(x)$ of a random variable $X$. Find the distribution function $G(y)$ of the random variable $Y=3X+2$. | Solution. By the definition of the distribution function, $G(y)=$ $=P(Y<y)$. Since the function $y=3 x+2$ is increasing, the inequality $Y<y$ holds if the inequality $X<x$ holds, therefore
$$
G(y)=P(Y<y)=P(X<x)=F(x)
$$
From the equation $y=3 x+2$, we express $x$:
$$
x=(y-2) / 3
$$
Substituting (**) into (*), we fin... | G(y)=F[(y-2)/3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,416 |
398. Given the distribution function $F(x)$ of a random variable $X$. Find the distribution function $G(y)$ of the random variable $Y=-(2 / 3) X+2$. | Solution. By the definition of the distribution function,
$$
G(y)=P(Y<x), \text{ therefore } G(y)=P(Y<x).
$$
Events $X<x$ and $X>x$ are complementary, so the sum of their probabilities is one: $P(X<x) + P(X>x) = 1$. Therefore,
$$
P(X>x)=1-P(X<x)=1-F(x) \text {; }
$$
hence,
$$
G(y)=1-F(x)
$$
From the equation $y=-... | G(y)=1-F[\frac{3(2-y)}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,417 |
402. Independent random variables $X$ and $Y$ are given by the probability density functions:
$f_{1}(x)=e^{-x} \quad(0 \leqslant x<\infty), \quad f_{2}(y)=(1 / 2) e^{-y / z} \quad(0 \leqslant y<\infty)$.
Find the composition of these laws, i.e., the probability density function of the random variable $Z=X+Y$. | Solution. Since the possible values of the arguments are non-negative, the formula
$$
g(z)=\int_{0}^{2} f_{1}(x) f_{2}(z-x) d x
$$
is applicable.
Therefore,
$$
g(2)=\int_{0}^{2} \mathrm{e}^{-x}\left[\frac{1}{2} \mathrm{e}^{-(2-x) / 2}\right] \mathrm{d} x
$$
After performing elementary transformations, we get
$$
g... | (z)=e^{-z/2}[1-\mathrm{e}^{-z/2}] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,419 |
404. Independent normally distributed random variables $X$ and $Y$ are given by the probability density functions:
$$
f_{1}(x)=(1 / \sqrt{2 \pi}) e^{-x^{2} / 2}, \quad f_{2}(y)=(1 / \sqrt{2 \pi}) e^{-y^{2} / 2}
$$
Prove that the composition of these laws, i.e., the probability density function of the random variable ... | Solution. We use the formula $g(z)=\int_{-\infty}^{\infty} f_{1}(x) f_{2}(z-x) \mathrm{d} x$. Then
$$
g(2)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \mathrm{e}^{-x^{2} / 2} \mathrm{e}^{-(2-x)^{2} / 2} \mathrm{~d} x
$$
Performing elementary calculations, we get
$$
g(z)=\frac{1}{2 \pi} \mathrm{e}^{-z^{2} / 2} \int_{-\in... | (z)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-z^{2}/4} | Algebra | proof | Yes | Yes | olympiads | false | 32,420 |
405. The densities of the distributions of independent uniformly distributed random variables $X$ and $Y$ are given: $f_{1}(x)=1 / 2$ in the interval $(0,2)$, outside this interval $f_{1}(x)=0$
$f_{2}(y)=1 / 2$ in the interval $(0,2)$, outside this interval $f_{2}(y)=0$
Find the distribution function and the density ... | Solution. According to the condition, the possible values of $X$ are determined by the inequality $0 < X < 4$. The function of distribution $G(z)$ is defined as follows:
- If $z \leq 0$, then $G(z)=0$.
- If $0 < z \leq 2$, then $G(z)=(1 / 8) S_{O A E D}=(1 / 8) z^{2}=z^{2} / 8$.
- If $2 < z \leq 4$, then $G(z)=(1 / 8)... | (z)={\begin{pmatrix}0&\text{for}z\leq0,\\z/4&\text{for}0<z\leq2,\\(4-z)/4&\text{for}2<z\leq4,\\0&\text{for}z>40\end{pmatrix}.} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,421 |
408. The probability distribution of a discrete two-dimensional random variable is given:
| $y$ | $x$ | | |
| :---: | :---: | :---: | :---: |
| | 3 | 10 | 12 |
| | | | |
| 4 | 0.17 | 0.13 | 0.25 |
| 5 | 0.10 | 0.30 | 0.05 |
Find the distribution laws of the components $X$ and $Y$. | Solution. By summing the probabilities "by columns", we obtain the probabilities of possible values of $X: p(3)=0.27, p(10)=0.43, p(12)=0.30$. We write the distribution law of the component $X$:
$$
\begin{array}{ccccc}
X & 3 & 10 & 12 \\
p & 0.27 & 0.43 & 0.30
\end{array}
$$
Check: $0.27+0.43+0.30=1$.
By summing the... | \begin{pmatrix}X&3&10&12\\p&0.27&0.43&0.30\end{pmatrix}\quad\begin{pmatrix}Y&4&5\\p&0.55&0.45\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,422 |
410. The distribution function of a two-dimensional random variable is given by
$$
F(x, y)=\left\{\begin{array}{c}
\sin x \cdot \sin y \text { for } 0 \leqslant x \leqslant \pi / 2,0 \leqslant y \leqslant \pi / 2 \\
0 \quad \text { for } x<0 \text { or } y<0
\end{array}\right.
$$
Find the probability that a random po... | Solution. We use the formula
$$
\begin{gathered}
P\left(x_{1}<X<x_{2}, y_{1}<Y<y_{2}\right)=\left[F\left(x_{2}, y_{2}\right)-F\left(x_{1}, y_{2}\right)\right]- \\
-\left[F\left(x_{2}, y_{1}\right)-F\left(x_{1}, y_{1}\right)\right] .
\end{gathered}
$$
Setting $x_{1}=0, x_{2}=\pi / 4, y_{1}=\pi / 6, y_{2}=\pi / 3$, we ... | \frac{(\sqrt{6}-\sqrt{2})}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,423 |
416. In the circle $x^{2}+y^{2} \leqslant R^{2}$, the two-dimensional probability density is $f(x, y)=C\left(R-\sqrt{x^{2}+y^{2}}\right)$; outside the circle $f(x, y)=0$. Find: a) the constant $C$; b) the probability that a random point $(X, Y)$ falls into the circle of radius $r=1$ centered at the origin, if $R=2$. | Solution. a) Using the second property of the two-dimensional probability density:
$$
\iint_{(D)} C\left(R-\sqrt{x^{2}+y^{2}}\right) \mathrm{d} x \mathrm{~d} y=1
$$
From this,
$$
C=1 / \int_{(D)}\left(R-\sqrt{\left.x^{2}+y^{2}\right)} \mathrm{d} x \mathrm{~d} y\right.
$$
By switching to polar coordinates, we get
$... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,424 |
421. A discrete two-dimensional random variable $(X, Y)$ is given:
| $Y$ | $X$ | | |
| :---: | :---: | :---: | :---: |
| | $x_{1}=2$ | $x_{2}=5$ | $x_{2}=8$ |
| | | | |
| $y_{1}=0.4$ | 0.15 | 0.30 | 0.35 |
| $y_{2}=0.8$ | 0.05 | 0.12 | 0.03 |
Find: a) the unconditional distribution laws of the components; b) t... | Solution. a) By summing the probabilities "by columns", we write the distribution law of $X$:
$$
\begin{array}{cccc}
X & 2 & 5 & 8 \\
p & 0.20 & 0.42 & 0.38
\end{array}
$$
By summing the probabilities "by rows", we find the distribution law of $y$:
$$
\begin{array}{lll}
y & 0.4 & 0.8 \\
p & 0.80 & 0.20
\end{array}
$... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,425 |
423. The joint density function of a continuous two-dimensional random variable $(X, Y)$ is given by
$$
f(x, y)=\frac{1}{\pi} \mathrm{e}^{-(1 / 2)\left(x^{2}+2 x y+3 y^{2}\right)}
$$
Find: a) the marginal density functions of the components; b) the conditional density functions of the components. | Solution. a) Let's find the density of the distribution of the component $X$:
$$
f_{1}(x)=\int_{-\infty}^{\infty} f(x, y) \mathrm{d} y=\frac{1}{\pi} \int_{-\infty}^{\infty} \mathrm{e}^{-(1 / 2)\left(x^{2}+2 x y+5 y^{2}\right)} \mathrm{d} y
$$
We factor out the multiplier $\mathrm{e}^{-x^{2} / 2}$, which does not depe... | \begin{aligned}&f_{1}(x)=\sqrt{\frac{2}{5\pi}}e^{-0.4x^{2}}\\&f_{2}(y)=\sqrt{\frac{2}{\pi}}e^{-2y^{2}}\\&\varphi(x\midy)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-0.5(x+y)^{2}}\\&\ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,426 |
430. The joint density function of a continuous two-dimensional random variable $(X, Y)$ is given by:
$$
f(x, y)= \begin{cases}4 x y e^{-x^{2}-y^{2}} & (x>0, y>0) \\ 0 & (x<0 \text { or } y<0) .\end{cases}
$$
Find: a) the expectations; b) the variances of the components $X$ and $Y$. | Solution. a) First, let's find the density of the distribution of the component $X$:
$$
f_{1}(x)=\int_{0}^{\infty} f(x, y) d y=4 x e^{-x^{2}} \int_{0}^{\infty} y e^{-y^{2}} d y=2 x e^{-x^{2}} \quad(x>0)
$$
Similarly, we get
$$
f_{2}(y)=2 y e^{-y^{2}} \quad(y>0)
$$
Let's find the expected value of the component $X$:... | M(X)=M(Y)=\frac{\sqrt{\pi}}{2},\quadD(X)=D(Y)=1-\frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,427 |
437. Prove that if the two-dimensional probability density function of the system of random variables $(X, Y)$ can be represented as the product of two functions, one of which depends only on $x$, and the other only on $y$, then the variables $X$ and $Y$ are independent. | Solution. By condition,
$$
f(x, y)=\varphi(x) \cdot \psi(y)
$$
Let's find the density distributions of the components:
$$
\begin{aligned}
& f_{1}(x)=\int_{-\infty}^{\infty} f(x, y) \mathrm{d} y=\varphi(x) \int_{-\infty}^{\infty} \psi(y) \mathrm{d} y \\
& f_{2}(y)=\int_{-\infty}^{\infty} f(x, y) \mathrm{d} x=\psi(y) ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,428 |
438. Prove that if $X$ and $Y$ are related by the linear dependence $Y=a X+b$, then the absolute value of the correlation coefficient is one. | Solution. By the definition of the correlation coefficient,
$$
r_{x y}=\mu_{x y} /\left(\sigma_{x} \sigma_{y}\right)
$$
where
$$
\mu_{x y}=M\{[X-M(X)][Y-M(Y)]\}
$$
Let's find the mathematical expectation of $Y$:
$$
M(Y)=M[a X+b]=a M(X)+b
$$
Substituting (**) into (*), after elementary transformations, we get
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,429 |
443. Construct a frequency polygon based on the given sample distribution:
$$
\begin{array}{rrrrr}
x_{i} & 1 & 4 & 5 & 7 \\
n_{i} & 20 & 10 & 14 & 6
\end{array}
$$ | Solution. We will plot the variants $x_{i}$ on the x-axis and the corresponding frequencies $n_{i}$ on the y-axis: connecting the points ( $x_{i}, n_{i}$ )
Fig. 12 with line segments, we obt... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 32,431 |
448. Construct a histogram of relative frequencies based on the given sample distribution:
| Interval Number $i$ | Partial Interval $\boldsymbol{x}_{i}-\boldsymbol{x}_{i+1}$ | Sum of Frequencies of Partial Interval $\qquad n_{i}$ |
| :---: | :---: | :---: |
| 1 2 3 | $0-2$ $2-4$ $4-6$ | 20 30 50 | | Solution. Let's find the relative frequencies:
$$
w_{1}=20 / 100=0.2, w_{2}=30 / 100=0.3, w_{3}=50 / 100=0.5
$$
Now, let's find the densities of the relative frequencies, considering that the length of the interval $h=2$:
$$
w_{1} / h=0.2 / 2=0.1, w_{2} / h=0.3 / 2=0.15, w_{3} / h=0.5 / 2=0.25 \text {. }
$$
We will... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 32,433 |
450. A sample of size $n=50$ has been drawn from the population:
| variant | $x_{i}$ | 2 | 5 | 7 | 10 |
| :--- | :--- | ---: | ---: | ---: | ---: |
| frequency | $n_{i}$ | 16 | 12 | 8 | 14 |
Find the unbiased estimate of the population mean. | The problem is solved. An unbiased estimate of the population mean is the sample mean
$$
\bar{x}_{\mathrm{B}}=\left(\sum n_{l} x_{i}\right) / n=(16.2+12.5+8.7+14 \cdot 10) / 50=5.76
$$ | 5.76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,434 |
452. The distribution of the initial sample variants of size $n$ is given:
$$
\begin{array}{lllll}
x_{i} & x_{1} & x_{2} & \cdots & x_{k} \\
n_{i} & n_{1} & n_{2} & \cdots & n_{k}
\end{array}
$$
Prove that
$$
\bar{x}_{\mathrm{B}}=C+\left(\sum n_{i} u_{i}\right) / n
$$
where the conditional variants $u_{i}=x_{i}-C$. | Solution. Since $u_{i}=x_{i}-C$, then $n_{i} u_{i}=n_{i}\left(x_{i}-C\right)$; summing the left and right sides of the equality over all values of $i$, we get $\sum n_{i} u_{i}=\sum n_{i}\left(x_{i}-C\right)$, or $\sum n_{i} u_{i}=\sum n_{i} x_{i}-C \sum n_{i}=\sum n_{i} x_{i}-C n$. Therefore,
$$
\sum n_{i} x_{i}=C n+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 32,435 |
453. Find the sample mean for the given distribution of sample size $n=10$ :
| $x_{i}$ | 1250 | 1270 | 1280 |
| :--- | :---: | :---: | :---: |
| $n_{i}$ | 2 | 5 | 3 | | Solution. The initial variants are large numbers, so we will switch to conditional variants $u_{i}=x_{i}-1270$. As a result, we will get the distribution of conditional variants:
$$
\begin{array}{rrrr}
u_{i} & -20 & 0 & 10 \\
n_{i} & 2 & 5 & 3
\end{array}
$$
We will find the required sample mean:
$$
\bar{x}_{\mathrm... | 1269 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,436 |
455. Based on a sample of size $n=41$, a biased estimate $D_{\text {v }}=3$ of the population variance was found. Find the unbiased estimate of the population variance. | Solution. The desired mixed estimate is equal to the corrected variance:
$$
s^{2}=\frac{n}{n-1} D_{B}=\frac{41}{40} \cdot 3=3.075
$$ | 3.075 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,437 |
457. As a result of five measurements of the length of a rod with one instrument (without systematic errors), the following results (in mm) were obtained: $92 ; 94 ; 103 ; 105 ; 106$. Find: a) the sample mean length of the rod; b) the sample and corrected variances of the instrument errors. | Solution. a) Let's find the sample mean:
$$
\bar{x}_{\mathrm{B}}=92+(0+2+11+13+14) / 5=92+8=100
$$
b) Let's find the sample variance:
$$
\begin{gathered}
D_{\mathrm{B}}=\frac{\sum\left(x_{i}-\bar{x}_{\mathrm{B}}\right)^{2}}{n}=\left[(92-100)^{2}+(94-100)^{2}+(103-100)^{2}\right] / 5+ \\
+\left[(105-100)^{2}+(106-100... | 100,34,42.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,438 |
460. Find the sample variance for the given sample distribution of size $n=10$:
$$
\begin{array}{cccc}
x_{i} & 186 & 192 & 194 \\
n_{i} & 2 & 5 & 3
\end{array}
$$ | Solution. The options are relatively large numbers, so we will switch to conditional variants $u_{i}=x_{i}-191$ (we subtract from the variant the number $C=191$, close to the sample mean). As a result, we will get the distribution of conditional variants:
$$
\begin{array}{lrrrr}
u_{i} & -5 & 1 & 3 \\
n_{i} & 2 & 5 & 3... | 8.04 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,439 |
463. Find the sample variance for the given sample distribution of size $n=10$:
$$
\begin{array}{cccc}
x_{i} & 0.01 & 0.04 & 0.08 \\
n_{i} & 5 & 3 & 2
\end{array}
$$ | Solution. To avoid operations with fractions, we will switch to conditional variants $u_{i}=100 x_{i}$. As a result, we will get the distribution
$$
\begin{array}{llll}
u_{i} & 1 & 4 & 8 \\
n_{i} & 5 & 3 & 2
\end{array}
$$
We will find the sample variance of the conditional variants:
$$
D_{\mathrm{B}}(u)=\left(\sum ... | 0.0007 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,440 |
466. Find the corrected sample variance for the given sample distribution $n=10$:
$$
\begin{array}{lccc}
x_{i} & 102 & 104 & 108 \\
n_{i} & 2 & 3 & 5
\end{array}
$$ | Solution. Let's switch to conditional variants $u_{i}=x_{i}-104$. In the end, we will get the distribution
$$
\begin{array}{lrll}
u_{i} & -2 & 0 & 4 \\
n_{i} & 2 & 3 & 5
\end{array}
$$
Let's find the corrected sample variance of the conditional variants:
$$
s_{u}^{2}=\frac{\sum n_{i} u_{i}^{2}-\left[\sum n_{i} u_{i}... | 6.93 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,441 |
468. Find the corrected sample variance for the given sample distribution of size $n=10$:
$$
\begin{array}{cccc}
x_{i} & 0.01 & 0.05 & 0.09 \\
n_{i} & 2 & 3 & 5
\end{array}
$$ | The problem is solved. To avoid operations with fractions, we will switch to conditional variants $u_{i}=100 x_{i}$. As a result, we get the distribution
$$
\begin{array}{llll}
u_{i} & 1 & 5 & 9 \\
n_{i} & 2 & 3 & 5
\end{array}
$$
We will find the corrected sample variance of the conditional variants
$$
s_{u}^{2}=\f... | 0.0085 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,442 |
480. Find the method of moments estimates for the unknown parameters $\alpha$ and $\beta$ of the gamma distribution from the sample $x_{1}, x_{2}, \ldots$, $x_{n}$, where the density is given by
$$
f(x)=\frac{1}{\beta^{\alpha+1 \Gamma}(\alpha+1)} x^{\alpha} e^{-x / \beta} \quad(\alpha>-1, \beta>0, x \geqslant 0)
$$ | Solution. To find two unknown parameters, it is necessary to have two equations; equate the initial theoretical moment of the first order $v_{1}$ to the initial empirical moment of the first order $M_{1}$ and the central theoretical moment of the second order $\mu_{3}$ to the central empirical moment of the second orde... | \alpha^{*}=(\overline{x_{\mathrm{B}}})^{2}/D_{\mathrm{B}}-1,\beta^{*}=D_{\mathrm{B}}/\overline{x_{\mathrm{B}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,444 |
481. The random variable $X$ (the water level in the river compared to the nominal level) is subject to a gamma distribution, the density of which is determined by the parameters $\alpha$ and $\beta$ $(\alpha>-1, \beta>0)$:
$$
f(x)=\frac{1}{\beta^{\alpha+1} \Gamma(\alpha+1)} x^{\alpha} \cdot \mathrm{e}^{-x / \beta} \q... | Solution. We use the point estimates of the parameters of the gamma distribution (see problem 480):
$$
\alpha^{*}=\left(\bar{x}_{\mathrm{B}}\right)^{2} / D_{\mathrm{B}}-1, \quad \beta^{*}=D_{\mathrm{B}} / \bar{x}_{\mathrm{B}}
$$
From the given distribution, we can easily find the sample mean and sample variance: $\ba... | \alpha^{*}=3.06,\beta^{*}=40.86 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,445 |
487. Find the method of moments estimates for the unknown parameters $\lambda_{1}$ and $\lambda_{2}$ of the "double Poisson distribution" based on the sample $x_{1}, x_{2}, \ldots, x_{n}$
$$
P\left(X=x_{i}\right)=\frac{1}{2} \cdot \frac{\lambda_{1}^{x_{i}}-\lambda_{1}}{x_{i}!}+\frac{1}{2} \cdot \frac{\lambda_{2}^{x_{i... | Solution. If the random variable $Z$ is distributed according to the Poisson law with parameter $\lambda$, then its theoretical initial moments of the first and second order are respectively (see problems 207, 227):
$$
\begin{aligned}
& v_{1}=M(Z)=\lambda \\
& v_{2}=M\left(Z^{2}\right)=\lambda+\lambda^{2}
\end{aligned... | \lambda_{1}=v_{1}-\sqrt{v_{2}-v_{1}-v_{1}^{2}},\lambda_{2}=v_{1}+\sqrt{v_{2}-v_{1}-v_{1}^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,446 |
489. Find the point estimate of the unknown parameter $p$ (the probability of an event occurring in a single trial) of the binomial distribution using the method of maximum likelihood:
$$
P_{m}\left(x_{i}\right)=C_{m}^{x_{i}} p^{x_{i}}(1-p)^{m-x_{i}}
$$
where $x_{i}$ is the number of occurrences of the event in the $... | Solution. Let's construct the likelihood function:
$$
L=p\left(x_{1} ; \theta\right) \cdot p\left(x_{2} ; \theta\right) \ldots p\left(x_{n} ; \theta\right)
$$
Considering that $\Theta=p$ and $P\left(X=x_{i}\right)=C_{m}^{x_{i}} p^{x_{i}}(1-p)^{m-x_{i}}$, we get
$L=\left[C_{m}^{x_{1}} p^{x_{1}}(1-p)^{m-x_{1}}\right] ... | p^{*}=(\sumx_{i})/(n) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,447 |
493. Find the point estimate of the unknown parameter $\lambda$ of the exponential distribution, with density $f(x)=\lambda \mathrm{e}^{-\lambda x}(x \geqslant 0)$, using the method of maximum likelihood based on the sample $x_{1}, x_{2}, \ldots, x_{n}$. | Solution. Let's form the likelihood function
$$
L=f\left(x_{1} ; \theta\right) \cdot f\left(x_{2} ; \theta\right) \ldots f\left(x_{n} ; \theta\right)
$$
considering that $\Theta=\lambda$ and, therefore, $f(x ; \theta)=f(x ; \lambda)=\lambda \mathrm{e}^{-\lambda x}$ :
$$
L=\left(\lambda \mathrm{e}^{-\lambda x_{1}}\ri... | \lambda^{*}=1/\bar{x}_{\text{B}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,448 |
501. Find the confidence interval for estimating the unknown expected value $a$ of a normally distributed feature $X$ of the population with a reliability of 0.95, if the population standard deviation $\sigma=5$, the sample mean $\bar{x}_{\text {v }}=14$ and the sample size $n=25$. | Solution. It is required to find the confidence interval
$$
\bar{x}_{\mathrm{B}}-t \frac{\sigma}{\sqrt{n}}<a<\bar{x}_{\mathrm{B}}+t \frac{\sigma}{\sqrt{n}}
$$
All quantities except $t$ are known. We will find $t$ from the relation $\Phi(t)=0.95 / 2=0.475$. From the table in Appendix 2, we find $t=1.96$. Substituting ... | 12.04<15.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,449 |
506. Find the minimum sample size for which with a reliability of 0.975, the accuracy of estimating the mean $a$ of the population using the sample mean is $\delta=0.3$, given that the standard deviation $\sigma=1.2$ of the normally distributed population is known. | Solution. We will use the formula that defines the accuracy of estimating the mathematical expectation of the general population by the sample mean: $\delta=t \sigma / \sqrt{n}$. From this,
$$
n=t^{2} \sigma^{2} / \delta^{2}
$$
According to the condition, $\gamma=0.975$; therefore, $\Phi(t)=0.975 / 2=0.4875$. From th... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,450 |
508. A sample of size $n=10$ is drawn from the general population:
$$
\begin{array}{lllllllll}
\text { value } & x_{i} & -2 & 1 & 2 & 3 & 4 & 5 \\
\text { frequency } & n_{i} & 2 & 2 & 2 & 2 & 2 & 1
\end{array}
$$
Estimate the mathematical expectation $a$ of the normally distributed characteristic of the general popu... | Solution. The sample mean and the "corrected" standard deviation will be found, respectively, by the formulas:
$$
\bar{x}_{\mathrm{B}}=\frac{\sum n_{i} x_{i}}{n}, s=\sqrt{\frac{\sum n_{i}\left(x_{i}-\bar{x}_{\mathrm{B}}\right)^{2}}{n-1}}
$$
Substituting the data from the problem into these formulas, we get $\overline... | 0.3<7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,451 |
510. According to the data of nine independent equi-precise measurements of a certain physical quantity, the arithmetic mean of the measurement results $\bar{x}_{\mathrm{v}}=30.1$ and the corrected mean square deviation $s=6$ have been found. Estimate the true value of the measured quantity using a confidence interval ... | Solution. The true value of the measured quantity is equal to its mathematical expectation a. Therefore, the problem reduces to estimating the mathematical expectation (with an unknown b) using a confidence interval
$$
\bar{x}_{B}-t_{\gamma} s / \sqrt{n}<a<\bar{x}_{B}+t_{\gamma} s / \sqrt{n}
$$
All quantities except ... | 25.38<34.82 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,452 |
514. Twelve measurements of a certain physical quantity were made using one instrument (without systematic error), and the "corrected" mean square deviation $s$ of the random errors of the measurements turned out to be 0.6. Find the accuracy of the instrument with a reliability of 0.99. It is assumed that the measureme... | Solution. The accuracy of the instrument is characterized by the root mean square deviation of random measurement errors. Therefore, the problem reduces to finding a confidence interval that covers $\sigma$ with a given confidence level $\gamma=0.99$:
$$
s(1-q)<\sigma<s(1+q)
$$
Given $\gamma=0.99$ and $n=12$, we find... | 0.06<\sigma<1.14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,454 |
516. Independent trials are conducted with the same but unknown probability $p$ of event $A$ occurring in each trial. Find the confidence interval for estimating the probability $p$ with a reliability of 0.95, if in 60 trials event $A$ occurred 15 times. | Solution. According to the condition, $n=60, m=15, \gamma=0.95$. We find the relative frequency of the occurrence of event $A: w=m / n=15 / 60=0.25$.
We find $t$ from the relation $\Phi(t)=\gamma / 2=0.95 / 2=0.475$. From the Laplace function table (see Appendix 2), we find $t=1.96$.
We find the boundaries of the des... | 0.16<p<0.37 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,455 |
518. An experimental slot machine has been created, which is supposed to ensure a win in one out of 100 coin tosses into the machine. To test the suitability of the machine, 400 trials were conducted, during which the win appeared 5 times. Find the confidence interval that covers the unknown probability of a win with a... | Solution. Let's find the relative frequency of winning: $w=m / n=5 / 400=0.0125$. Let's find $t$ from the relation $\Phi(t)=$ $=\gamma / 2=0.999 / 2=0.4995$. From the table of the Lannace function (see Appendix 2), we find $t=3.3$.
Considering that $n=400$ is large, we will use approximate formulas to find the boundar... | 0<p<0.0308 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,456 |
523. Find the sample mean and sample variance using the product method for the given sample distribution of size $n=100$:
| variant | $x_{i}$ | 12 | 14 | 16 | 18 | 20 | 22 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| frequency | $n_{i}$ | 5 | 15 | 50 | 16 | 10 | 4 | | Solution. We will form Calculation Table 1; for this:
1) we will write the variants in the first column;
2) we will write the frequencies in the second column; the sum of the frequencies (100) will be placed in the bottom cell of the column;
3) as the false zero $C$, we will choose the variant (16), which has the high... | \bar{x}_{\mathrm{B}}= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,457 |
525. Find the sample mean and sample variance using the method of products for the given sample distribution of size \( n=100 \):
| \( x_{i} \) | 2 | 3 | 7 | 9 | 11 | 12.5 | 16 | 18 | 23 | 25 | 26 |
| ---: | ---: | ---: | ---: | ---: | ---: | :---: | ---: | ---: | ---: | ---: | ---: |
| \( n_{i} \) | 3 | 5 | 10 | 6 | ... | Solution. Let's divide the interval $2-26$ into four sub-intervals of length $h=6: 2-8 ; 8-14 ; 14-20 ; 20-26$. Taking the midpoints of the sub-intervals as new variants $y_{i}$, we obtain equidistant variants: $y_{1}=5, y_{2}=11, y_{3}=17, y_{4}=23$.
As the frequency $n_{1}$ of the variant $y_{1}=5$, we take the sum ... | 42.14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,458 |
529. Find the sample mean and sample variance using the summation method for the given sample distribution of size $n=100$:
| variant | $x_{i}$ | 48 | 52 | 56 | 60 | 64 | 68 | 72 | 76 | 80 | 84 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| frequency | $n_{i}$ | 2 | 4 | 6 | 8... | Solution. We will form Calculation Table 2 as follows:
1) write the variants in the first column;
2) write the frequencies in the second column; place the sum of the frequencies (100) in the bottom cell of the column;
3) as the false zero $C$, choose the variant (68), which has the highest frequency (as $C$, you can t... | 68.12,64.78 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,459 |
531. Find the skewness and 9-excess by the method of products for the given sample distribution of size $n=100$:
$$
\begin{array}{lrrrrrrr}
\text { variant } & x_{i} & 12 & 14 & 16 & 18 & 20 & 22 \\
\text { frequency } & n_{i} & 5 & 15 & 50 & 16 & 10 & 4
\end{array}
$$ | Solution. We will use the method of products. We will compile the calculation table 3. In § 1 of this chapter, when solving problem 523, it was already indicated how to fill in columns 1-5 of the calculation table, so we will limit ourselves to brief explanations.
To fill in column 6, it is convenient to multiply the ... | a_{}=0.47,e_{k}=0.36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,460 |
533. Find the skewness and kurtosis by the method of sums for the given sample distribution of size \( n=100 \):
\[
\begin{array}{rrrrrrrrrrr}
x_{i} & 48 & 52 & 56 & 60 & 64 & 68 & 72 & 76 & 80 & 84 \\
n_{i} & 2 & 4 & 6 & 8 & 12 & 30 & 18 & 8 & 7 & 5
\end{array}
\] | Solution. We will use the method of sums, for this purpose we will compose the calculation table 4. In § 2 of this chapter, when solving problem 529, it was already indicated how to fill in columns $1-4$ of the calculation table, so we will limit ourselves to brief explanations.
To fill in column 5, we will write zero... | a_{}=-0.25,\;e_{k}=26.97 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,461 |
540. The knowledge of ten students was tested in two tests: $A$ and $B$. The scores on a hundred-point scale were as follows (the first row shows the scores for test $A$, and the second row shows the scores for test $B$):
| 95 | 90 | 86 | 84 | 75 | 70 | 62 | 60 | 57 | 50 | (*) |
| :--- | :--- | :--- | :--- | :--- | :-... | The solution is as follows. Assign ranks $x_{i}$ to the scores of test $A$. These scores are arranged in descending order, so their ranks $x_{i}$ are equal to their ordinal numbers:
$$
\begin{array}{lllllrrrrrrrr}
\text { ranks } x_{i} \\
\text { scores of } & & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\text { test }... | 0.64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,464 |
546. Two controllers $A$ and $B$ arranged samples of products made by nine masters in the order of decreasing quality (in parentheses are the ordinal numbers of products of the same quality):
(A)
$\begin{array}{llll}\text { (A) } & 1 & 2 \\ \text { (B) } & 2 & 1\end{array}$
$\left.\begin{array}{ccc}(3, & 4 & 5 \\ 4 & 3... | The solution is as follows. Given that the ranks of items of the same quality are equal to the arithmetic mean of the ordinal numbers of the items: $(3+4+5) / 3=4, \quad(6+7+8+9) / 4=7.5, \quad(6+7) / 2=6.5$, we can write the sequences of ranks assigned to the items by the controllers:
$$
\begin{array}{cccccccccc}
x_{... | 0.93 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,465 |
548. The knowledge of 10 students was tested in two tests: $A$ and $B$. The scores on a 100-point scale were as follows (the first row shows the scores for test $A$, and the second row shows the scores for test $B$):
| 95 | 90 | 86 | 84 | 75 | 70 | 62 | 60 | 57 | 50 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |... | Solution. When solving problem 540, the condition of which coincides with the condition of the current problem, two rank sequences were obtained (in the first row, the ranks by test $A$ are given, in the second row - by test $B$):
$$
\begin{array}{rrrrrrrrrrr}
x_{i} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
y_{i} & ... | 0.47 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,466 |
554. From two independent samples, with sizes $n_{1}=11$ and $n_{2}=14$, extracted from normal general populations $X$ and $Y$, the corrected sample variances $s_{X}^{2}=0.76$ and $s_{Y}^{2}=0.38$ were found. At the significance level $\alpha=0.05$, test the null hypothesis $H_{0}: D(X)=$ $=D(Y)$ about the equality of ... | Solution. Let's find the ratio of the larger corrected variance to the smaller one:
$$
F_{\text {obs }}=0.76 / 0.38=2
$$
According to the condition, the alternative hypothesis is $D(X)>D(Y)$, so the critical region is right-tailed.
From the table in Appendix 7, at the significance level $\alpha=0.05$ and degrees of ... | F_{\text{obs}}<F_{\text{crit}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,467 |
556. For two independent samples, the sizes of which are $n_{1}=14$ and $n_{2}=10$, extracted from normal general populations $X$ and $Y$, the corrected sample variances are $s_{X}^{2}=0.84$ and $s_{Y}^{2}=2.52$. At the significance level $\alpha=0.1$, test the null hypothesis $H_{0}: D(X)=$ $=D(Y)$ about the equality ... | Solution. Let's find the ratio of the larger corrected variance to the smaller one:
$$
F_{\text {obs }}=2.52 / 0.84=3 .
$$
According to the condition, the alternative hypothesis is $D(X) \neq D(Y)$, so the critical region is two-tailed. According to rule 2, when finding the critical point, the significance level shou... | F_{\text{obs}}>F_{\text{crit}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,468 |
558. Two methods were used to measure the same physical quantity. The following results were obtained:
a) in the first case $x_{1}=9.6 ; \quad x_{2}=10.0 ; x_{3}=9.8$; $x_{4}=10.2 ; x_{5}=10.6$
b) in the second case $y_{1}=10.4 ; \quad y_{2}=9.7 ; \quad y_{3}=10.0$; $y_{4}=10.3$.
Can we consider that both methods pr... | Solution. We will judge the accuracy of the methods by the magnitudes of the variances. Thus, the null hypothesis is $H_{0}: D(X) = D(Y)$. As a competing hypothesis, we will take $H_{1}: D(X) \neq D(Y)$.
Let's find the sample variances. For simplification of calculations, we will transition to conditional variants:
$... | F_{\text{obs}}<F_{\text{crit}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,469 |
560. A sample of size $n=21$ has been drawn from a normal population, and the corrected sample variance $s^{2}=16.2$ has been calculated. It is required to test the null hypothesis $H_{0}$ : $\sigma^{2}=\sigma_{0}^{2}=15$ at a significance level of 0.01, using the alternative hypothesis $H_{1}: \sigma_{0}^{2}>15$. | Solution. Let's find the observed value of the criterion:
$$
\chi_{\text {obs }}^{2}=\frac{(n-1) s^{2}}{\sigma_{0}^{2}}=\frac{(21-1) \cdot 16.2}{15}=21.6 .
$$
According to the condition, the alternative hypothesis is $\sigma^{2}>15$, so the critical region is right-sided (Rule 1). From the table in Appendix 5, at a s... | \chi_{\text{obs}}^{2}<\chi_{\text{crit}}^{2} | Other | math-word-problem | Yes | Yes | olympiads | false | 32,470 |
563. The accuracy of an automatic machine is checked by the variance of the controlled size of the products, which should not exceed $\sigma_{0}^{2}=0.1$. A sample of 25 randomly selected products was taken, and the following measurement results were obtained:
controlled size
$$
\begin{array}{lllllllll}
\text { of th... | Solution. Null hypothesis $H_{0}: \sigma^{2}=\sigma_{0}^{2}=0.1$. We will take as the alternative hypothesis $H_{1}: \sigma^{2}>0.1$.
Let's find the corrected sample variance. For simplification of the calculation, we will transition to conditional variants. Taking into account that the sample mean is approximately eq... | \chi_{\text{obs}}^{2}>\chi_{\text{crit}}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,471 |
565. A batch of products is accepted if the variance of the controlled size does not significantly exceed 0.2. The corrected sample variance, found from a sample of size \( n=121 \), turned out to be \( s_{x}^{2}=0.3 \). Can the batch be accepted at a significance level of 0.01? | Solution. Null hypothesis $H_{0}: \sigma^{2}=\sigma_{0}^{2}=0.2$. Alternative hypothesis $H_{1}: \sigma^{2}>0.2$.
Let's find the observed value of the criterion:
$$
\chi_{\text {Obs }}^{2}=(n-1) \cdot s_{X}^{2} / \sigma_{0}^{2}=120 \cdot 0.3 / 0.2=180 .
$$
The alternative hypothesis is $\sigma^{2}>0.2$, hence the cr... | Thebatchcannotbeaccepted | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,472 |
567. From two independent samples, the sizes of which are $n=40$ and $m=50$, extracted from normal populations, the sample means are found to be $\bar{x}=130$ and $\bar{y}=140$. The population variances are known: $D(X)=80$, $D(Y)=100$. It is required to test the null hypothesis $H_{0}: M(X)=M(Y)$ against the alternati... | Solution. Let's find the observed value of the criterion:
$$
z_{\text {obs }}=\frac{\bar{x}-\bar{y}}{\sqrt{D(X) / n+D(Y) / m}}=\frac{130-140}{\sqrt{80 / 40+100 / 50}}=-5
$$
According to the condition, the alternative hypothesis is $M(X) \neq M(Y)$, so the critical region is two-sided.
Let's find the right critical p... | |Z_{\text{obs}}| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,473 |
572. From two batches of products manufactured on two identically configured machines, small samples of sizes $n=10$ and $m=12$ were extracted. The following results were obtained:
controlled size
of products from the first machine $x_{i} \quad 3.4 \quad 3.5 \quad 3.7 \quad 3.9$ frequency (number of products) $n_{i} ... | Solution. Using the formulas
$$
\bar{x}=\left(\sum n_{i} x_{i}\right) / n \quad \text { and } \quad \bar{y}=\left(\sum m_{i} y_{i}\right) / m
$$
we find the sample means: $\bar{x}=3.6, \bar{y}=3.5$.
To simplify the calculations of the corrected variances, we will use the conditional variables $u_{i}=10 x_{i}-36, v_{... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,474 |
574. From a normal general population with a known standard deviation $\sigma=5.2$, a sample of size $n=100$ was extracted, and the sample mean $\bar{x}=27.56$ was found. It is required to test the null hypothesis $H_{0}: a=$ $=a_{0}=26$ against the alternative hypothesis $H_{1}: a \neq 26$ at a significance level of 0... | Solution. Let's find the observed value of the criterion:
$$
U_{\text {obs } 6 \pi}=\left(\bar{x}-a_{0}\right) \cdot \sqrt{n} / \sigma+(27.56-26) \cdot \sqrt{100} / 5.2=3 .
$$
According to the condition, the alternative hypothesis is $a \neq a_{0}$, so the critical region is two-sided.
Let's find the critical point ... | 3>1.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,475 |
576. a) From a sample of size $n$, extracted from a normal general population with a known standard deviation $\sigma$, the sample mean $\bar{x}$ is found. At a significance level $\alpha$, it is required: 1) to find the critical region if the null hypothesis $H_{0}$: $a=a_{0}$ about the equality of the general mean $a... | Solution. 1) We use the formula
$$
1-\beta=0.5-\Phi\left(u_{\mathrm{kp}}-\lambda\right) .
$$
By rule 2, we find the critical point of the right-sided critical region $u_{\text {kp}}=1.65$.
We calculate $\lambda$, considering that, according to the condition, $a_{1}=3, a_{0}=2, n=16$, $\sigma=4:$
$$
\lambda=\left(a_... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,476 |
577. a) From a sample of size $n$, extracted from a normal general population with a known standard deviation $\sigma$, the sample mean $\bar{x}$ has been found. At a significance level $\alpha$, it is required to find the power function of the test criterion for testing the null hypothesis $H_{0}: a=a_{0}$ about the e... | Solution. We use the formula
$$
1-\beta=1-\left[\Phi\left(u_{\mathbf{k} p}-\lambda\right)+\Phi\left(u_{\mathbf{k} p}+\lambda\right)\right] .
$$
By rule 1, we find the critical point $u_{\text {cr }}=1.96$.
We calculate $\lambda$, considering that, according to the condition, $a_{1}=24, a_{0}=20$, $n=16, \sigma=5:$
... | 0.8925 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,477 |
581. Two instruments measured six parts in the same order and the following measurement results were obtained (in hundredths of a millimeter):
\[
\begin{aligned}
& x_{1}=2, x_{2}=3, x_{3}=5, x_{4}=6, x_{5}=8, x_{6}=10 ; \\
& y_{1}=10, y_{2}=3, y_{3}=6, y_{4}=1, y_{5}=7, y_{6}=4
\end{aligned}
\]
At a significance leve... | Solution. Let's find the differences $d_{i}=x_{i}-y_{i}$; subtracting the numbers of the second row from the numbers of the first, we get:
$$
d_{1}=-8, d_{2}=0, d_{3}=-1, d_{4}=5, d_{5}=1, d_{6}=6
$$
Let's find the sample mean, considering that $\sum d_{i}=3$: $\bar{d}=$ $=3 / 6=0.5$.
Let's find the "corrected" stan... | 0.24<2.57 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,478 |
588. A batch of products is accepted if the probability that a product will be defective does not exceed 0.02. Among 480 randomly selected products, 12 were found to be defective. Can the batch be accepted? | Solution. The null hypothesis $H_{0}$ is $p=p_{0}=0.02$. Let's find the relative frequency of defects:
$$
m / n=12 / 480=0.025
$$
We will take the alternative hypothesis $H_{1}: p>0.02$ and the significance level $\alpha=0.05$.
Let's find the observed value of the criterion:
$$
U_{\text {obs}}=\frac{\left(m / n-p_{... | Thebatchcanbeaccepted | Other | math-word-problem | Yes | Yes | olympiads | false | 32,480 |
592. For three independent samples with sizes $n_{1}=9, n_{2}=13$ and $n_{3}=15$, extracted from normal general populations, the corrected sample variances were found to be 3.2, 3.8, and 6.3, respectively. It is required to test the null hypothesis of homogeneity of variances at a significance level of 0.05. | Solution. We will compile the calculation table 10 (column 8 will not be filled in for now, as it is not yet known whether we will need to calculate $C$).
Using the calculation table, we find:
$$
\begin{gathered}
\overline{s^{2}}=\left(\sum k_{i} s_{i}^{2}\right) / k=159.4 / 34=4.688 ; \quad \lg \overline{s^{2}}=0.67... | 1.43 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,481 |
599. For four independent samples of the same size \( n=17 \), extracted from normal general populations, the corrected sample variances are: 0.21; 0.25; 0.34; 0.40.
Required: a) at a significance level of 0.05, test the null hypothesis of homogeneity of variances (critical region - right-tailed); b) estimate the popu... | Solution. a) Let's find the observed value of the Cochran criterion - the ratio of the maximum corrected variance to the sum of all variances:
$$
G_{\text {obs } 6 \pi}=0.40 /(0.21+0.25+0.34+0.40)=\frac{1}{3} .
$$
We will find the critical point $G_{\text {crit }}(0.05 ; 16 ; 4)=0.4366$ in the table of critical value... | 0.3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,482 |
606. During the shift, 15 elements of device 1, consisting of 800 elements, and 25 elements of device 2, consisting of 1000 elements, failed. At the significance level $\alpha=0.05$, test the null hypothesis $H_{0}: p_{1}=p_{2}=p$ about the equality of the failure probabilities of the elements of both devices against t... | Solution. According to the condition, the alternative hypothesis has the form $\dot{p}_{1} \neq p_{2}$, so the critical region is two-sided. Let's find the observed value of the criterion:
$$
U_{\text {obs } \sigma}=\frac{m_{1} \cdot n_{1}-m_{2} / n_{2}}{\sqrt{\frac{m_{1}+m_{2}}{n_{1}+n_{2}} \cdot\left(1-\frac{m_{1}+m... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 32,483 |
610. Based on a sample of size $n=100$, extracted from a two-dimensional normal population ( $X, Y$ ), the sample correlation coefficient $r_{\mathrm{B}}=0.2$ has been found. It is required to test the null hypothesis of the population correlation coefficient being zero against the alternative hypothesis $H_{1}: r_{\ma... | Solution. Let's find the observed (empirical) value of the criterion:
$$
T_{\text {na } 6 \pi}=r_{\mathrm{B}} \sqrt{n-2} / \sqrt{1-r_{\mathrm{B}}^{2}}=0.2 \sqrt{100-2} / \sqrt{1-0.2^{\mathrm{z}}}=2.02
$$
According to the condition, the alternative hypothesis is $r_{\mathbf{r}} \neq 0$, so the critical region is two-t... | 2.02>1.99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,484 |
613. Based on a sample of size $n=100$, extracted from a two-dimensional normal population ($X, Y$), a correlation table 12 has been compiled.
Table 12
| $Y$ | $x$ | | | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 10 | 15 | 20 | 25 | 30 | 35 | $n_{y}$ |
| 35 | 5 | 1 | - | - | ... | Solution. a) To simplify the calculations, we switch to conditional variants
$$
u_{i}=\left(x_{i}-C_{1}\right) / h_{1}, \quad v_{i}=\left(y_{i}-C_{2}\right) / h_{2}
$$
where $C_{1}$ and $C_{2}$ are false zeros (it is advantageous to take a variant located approximately in the middle of the variation series; in this c... | 0.817 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,485 |
617. In problem 540, for a sample size of $n=10$, the sample Spearman rank correlation coefficient $\rho_{\mathrm{e}}=0.64$ was calculated between the test scores of students in two tests. At a significance level of 0.01, test the null hypothesis that the population Spearman rank correlation coefficient is zero. In oth... | Solution. We will find the critical point of the two-sided critical region of the Student's distribution at the significance level $\alpha=0.01$ and the number of degrees of freedom $k=n-2=10-2=8$ (see Appendix 6$): t_{\text {cr }}(0.01 ; 8)=3.36$.
We will find the critical point:
$$
T_{\mathrm{kp}}=t_{\mathrm{kp}}(\... | 0.92 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,486 |
623. In problem 548, for a sample size of $n=10$, the sample coefficient of rank correlation Kendall $\tau_{\text {v }}=0.47$ was calculated between the assessment scores of students on two tests. At a significance level of 0.05, test the null hypothesis that the population coefficient of rank correlation Kendall is ze... | Solution. Let's find the critical point $z_{\text {cr }}$:
$$
\Phi\left(z_{\mathbf{k p}}\right)=(1-\alpha) / 2=(1-0.05) / 2=0.475
$$
From the Laplace table (see Appendix 2), we find $z_{\text {cr }}=1.96$.
Let's find the critical point:
$$
T_{\mathrm{kp}}=z_{\mathrm{kp}} \sqrt{\frac{2(2 n+5)}{9 n(n-1)}}=1.96 \sqrt{... | \tau_{v}<T_{ | Other | math-word-problem | Yes | Yes | olympiads | false | 32,487 |
627. At a significance level of 0.01, test the null hypothesis $H_{0}: F_{1}(x)=F_{2}(x)$ about the homogeneity of two samples, the sizes of which are $n_{1}=6, n_{2}=7$ (the first row contains the variants of the first sample; the second row contains the variants of the second sample):
$$
\begin{array}{rrrrrrrr}
x_{i... | Solution. The competing hypothesis has $F_{1}(x) \neq F_{3}(x)$, so the critical region is two-sided. Arrange the variants of both samples in the form of a single variational series and number them:
| ordinal | | | | | | | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :---... | 24<41<60 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,488 |
3. At a significance level of 0.05, test the null hypothesis of homogeneity of two samples of sizes: $n_{1}=40$ and $n_{2}=50$ under the alternative hypothesis $H_{1}: F_{1}(x) \neq F_{2}(x)$, if it is known that in the combined variational series, composed of the variants of both samples, the sum of the ordinal number... | Solution. According to the condition, the competing hypothesis has $F_{1}(x)=F_{2}(x)$, so the critical region is two-sided.
Let's find $\boldsymbol{z}_{\text {kp }}$ using the equality
$$
\Phi\left(z_{\mathrm{kp}}\right)=(1-\alpha) / 2=(1-0.05) / 2=0.475
$$
Using the Laplace function table (see Appendix 2), we find... | 1578<1800<2062 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,489 |
637. Using the Pearson criterion, at a significance level of 0.01, determine whether the discrepancy between the empirical frequencies $n_{i}$ and the theoretical frequencies $n_{l}^{\prime}$, which are calculated based on the hypothesis of a normal distribution of the general population $X$, is random or significant:
... | Solution. Let's find the observed value of the Pearson criterion: $\chi_{\text {obs }}^{2}=\sum\left(n_{i}-n_{i}^{\prime}\right)^{2} / n_{i}^{\prime}$. We will create the calculation table 19.
From table 19, we find the observed value of the criterion: $\chi_{1 \text { obs }}^{2}=3.061$.
Using the table of critical p... | 3.061 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,491 |
645. From the general population $X$, a sample of size $n=50$ has been drawn, not grouped into intervals (in the first row, the variants are indicated, and in the second row, the corresponding frequencies):
| $c_{i}$ 1,40 $i$ 1 | 1,52 1 | 1,63 1 | 1,69 1 | | 73 2 | 1,78 | 1,89 1 | 1,92 1 | 1,95 1 |
| :---: | ... | Solution. 1. Let's form the calculation table No. 30. | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 32,494 |
647. Why, when checking the hypothesis of exponential distribution of the population by the Pearson criterion, is the number of degrees of freedom determined by the equation $k=s-2$, where $s$ is the number of sample intervals? | Solution. When using the Pearson criterion, the number of degrees of freedom is $k=s-1-r$, where $r$ is the number of parameters estimated from the sample. The exponential distribution is defined by one parameter $\lambda$. Since this parameter is estimated from the sample, $r=1$, and therefore, the number of degrees o... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,495 |
652. $n=100$ experiments were conducted. Each experiment consisted of $N=10$ trials, in each of which the probability $p$ of event $A$ occurring was 0.3. As a result, the following empirical distribution was obtained (in the first row, the number $x_{i}$ of occurrences of event $A$ in one experiment is indicated; in th... | Solution. 1. By the Bernoulli formula
$$
P_{i}=P_{N}(i)=C_{N}^{i} p^{i} q^{N-i}
$$
we find the probability $P_{i}(i=0,1,2,3,4,5)$ that event $A$ will occur exactly $i$ times in $N=10$ trials.
Noting that $p=0.3, q=1-0.3=0.7$, we get:
$$
P_{0}=P_{10}(0)=0.7^{10}=0.0282 ; P_{1}=P_{10}(1)=10 \cdot 0.3 \cdot 0.7^{9}=0.... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 32,496 |
5. Why are the parameters $a$ and $b$ of a uniformly distributed random variable $X$ estimated by the formulas
$$
a^{*}=\bar{x}_{\mathrm{B}}-\sqrt{3} \sigma_{\mathrm{B}}, \quad b^{*}=\bar{x}_{\mathrm{B}}+\sqrt{3} \sigma_{\mathrm{B}} ?
$$ | Solution. It is known that the sample mean $\bar{x}_{\mathrm{B}}$ and the sample standard deviation $\sigma_{\mathbf{z}}$ can be used as estimates for the expected value and standard deviation of a random variable $X$, respectively.
It is also known (see Chapter VI, problems 313, 315) that for a uniform distribution, ... | ^{*}=\bar{x}_{\mathrm{B}}-\sqrt{3}\sigma_{\mathrm{B}},\quadb^{*}=\bar{x}_{\mathrm{B}}+\sqrt{3}\sigma_{\mathrm{B}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,497 |
657. Why, when testing the hypothesis of a uniform distribution of the population $X$ using the Pearson's criterion, is the number of degrees of freedom determined by the equation $k=s-3$, where $s$ is the number of intervals in the sample? | Solution. When using the Pearson criterion, the number of degrees of freedom $k=s-1-r$, where $r$ is the number of parameters estimated from the sample. The uniform distribution is defined by two parameters $a$ and $b$. Since these two parameters are estimated from the sample, $r=2$, and therefore, the number of degree... | -3 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,498 |
6. The quality control department inspected $n=200$ batches of identical products and obtained the following empirical distribution (the first row indicates the number $x_{i}$ of non-standard items in one batch; the second row indicates the frequency $n_{i}$, i.e., the number of batches containing $x_{i}$ non-standard ... | Solution. 1. Find the sample mean:
$$
\bar{x}_{\mathrm{B}}=\left(\sum n_{i} x_{i}\right) / n=(116 \cdot 0+56 \cdot 1+22 \cdot 2+4 \cdot 3+2 \cdot 4) / 200=0.6
$$ | 0.6 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,500 |
668. Four tests were conducted at each of the three levels of factor $F$. Using the method of variance analysis, test the null hypothesis of equality of group means at a significance level of 0.05. It is assumed that the samples were drawn from normal populations with equal variances. The test results are given in Tabl... | Solution. To simplify the calculation, we subtract the overall mean $\bar{x}=29$ from each observed value $x_{i}$, i.e., we transition to reduced values: $y_{i j}=x_{i j}-29$. For example, $y_{11}=x_{11}-29=38-29=9 ; y_{21}=x_{21}-29=36-29=7$ and so on.
We will construct the calculation table 45.
Table
45
| Test Nu... | F_{\text{obs}}>F_{\text{crit}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,501 |
683. The probabilities of three events $A_{1}, A_{2}, A_{3}$, forming a complete group, are given: $p_{1}=P\left(A_{1}\right)=0.22, p_{2}=$ $=P\left(A_{2}\right)=0.31, p_{3}=P\left(A_{3}\right)=0.47$. Conduct five trials, in each of which one of the three considered events occurs. | Solution. According to the rule of this paragraph, we need to simulate a discrete random variable $\boldsymbol{X}$ with the following distribution:
$$
\begin{array}{cccc}
X & 1 & 2 & 3 \\
p & 0.22 & 0.31 & 0.47
\end{array}
$$
According to rule $\S 1$, we divide the interval (0,1) into three sub-intervals: $\Delta_{1}... | A_{3},A_{1},A_{3},A_{1},A_{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,504 |
689. Find the explicit formula for the generation of a continuous random variable $X$, uniformly distributed in the interval ( $a, b$ ), given its distribution function $F(x)=(x-a) /(b-a)(a<x<b)$. | Solution. According to Rule 1, we equate the given distribution function to a random number $r_{i}$ :
$$
\left(x_{i}-a\right) /(b-a)=r_{i}
$$
Solving this equation for $x_{i}$, we obtain the explicit formula for generating possible values of $X: x_{i}=(b-a) r_{i}+a$. | x_{i}=(b-)r_{i}+ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,505 |
692. Find an explicit formula for the simulation of a continuous random variable given by the probability density function $f(x)=b /(1+a x)^{2}$ in the interval $[0,1 /(b-a)]$; outside this interval $f(x)=0$. | Solution. Using rule 2, we write the equation
$$
b \int_{0}^{x_{i}} 1 /(1+a x)^{2} \mathrm{~d} x=r_{i}
$$
Solving this equation for $x_{i}$, we finally obtain
$$
x_{i}=r_{i} /\left(b-a r_{i}\right)
$$ | x_{i}=r_{i}/(b-r_{i}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,506 |
704. Find by the method of superposition explicit formulas for the generation of a continuous random variable $X$, given by the distribution function $F(x)=1-(1 / 3)\left(2 \mathrm{e}^{-2 x}+\right.$ $\left.+\mathrm{e}^{-3 x}\right)(0<x<\infty)$. | Solution. According to rule 3, we represent the given function as
$$
F(x)=(1 / 3)\left(1-e^{-3 x}\right)+(2 / 3)\left(1-e^{-2 x}\right)
$$
The functions enclosed in parentheses are distribution functions of the exponential law, so we can assume: $F_{1}(x)=1-\mathrm{e}^{-3 x}$, $F_{2}(x)=1-\mathrm{e}^{-2 x}, C_{1}=1 /... | \begin{aligned}&(-\lnr_{2})/3,\text{if}r_{1}<1/3,\\&(-\lnr_{2})/2,\text{if}r_{1}\geqslant1/30\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,507 |
710. Generate four possible values of a normal random variable with parameters: a) $a=0, \sigma=1$; b) $a=2, \sigma=3$. | Solution. a) In accordance with the rule, let's determine the possible value of $x_{1}$ of the normal random variable $X$ with parameters $a=0$ and $\sigma=1$ using the formula
$$
x_{1}=\sum_{i=1}^{12} r_{j}-6=S_{1}-6
$$
Select the first 12 random numbers from the second row of the table in Appendix 9: 0.37; 0.54; 0.... | z_{1}=2.18,z_{2}=-1.3,z_{3}=-2.56,z_{4}=4.49 | Other | math-word-problem | Yes | Yes | olympiads | false | 32,508 |
714. A discrete two-dimensional random variable ( $X, Y$ ), the components of which are independent, is given by the distribution law:
| $Y$ | $X$ | | |
| :---: | :---: | :---: | :---: |
| | $x_{1}$ | $x_{2}$ | $x_{3}$ |
| | | | |
| $y_{1}$ | 0.18 | 0.30 | 0.12 |
| $y_{2}$ | 0.12 | 0.20 | 0.08 |
Simulate the r... | Solution. Let's find the distribution law of the component $X$:
$$
\begin{aligned}
& p_{1}=P\left(X=x_{1}\right)=0.18+0.12=0.30 \\
& p_{2}=P\left(X=x_{2}\right)=0.30+0.20=0.50, \\
& p_{3}=P\left(X=x_{3}\right)=0.12+0.08=0.20 .
\end{aligned}
$$
Thus, the required distribution law is
$$
\begin{array}{cccc}
X & x_{1} &... | Other | math-word-problem | Yes | Yes | olympiads | false | 32,509 | |
5. Find the explicit formulas for the distribution of the continuous two-dimensional random variable ( $X, Y$ ), given by the probability density function $f(x, y)=(3 / 4) x y^{2}$ in the region bounded by the lines $x=0, y=0, x=1, y=2$. | Solution. Components $X$ and $Y$ are independent, as the joint probability density $f(x, y)=(3 / 4) x y^{2}$ can be represented as the product of two functions, one depending only on $x$, and the other only on $y$.
Let's find the density of the distribution of component $X$:
$$
f_{1}(x)=\int_{0}^{2} f(x, y) \mathrm{d... | x_{i}=\sqrt{r_{i}},\quady_{i}=2\sqrt[3]{r_{i}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,510 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.