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1. Problem 11

1.1. Question

There are \(841\) pupils in some school. Probability, that randomly chosen pupil has at least one D is equal to \(\frac{1}{12}\). Calculate probability \(p\), that number of pupils that have at least one D differs from \(\frac{841}{12}\) at most by 20. Consider absolute difference!

  1. \(p \geq 0.99371\)
  2. \(p \approx 0.98741\)
  3. \(p \approx 0.78741\)
  4. \(p < 0.89371\)
  5. \(p \leq 0.88741\)

1.2. Answer

2 is true; the rest is false; 5 is impossible

If \(P(X=k)\) denotes the probability that exactly k pupils have at least one D, then \(P(X=k) = \binom{841}{k}(\frac{1}{12})^k (\frac{11}{12})^{841-k}\)

\(P(|X-\frac{841}{12}|<20) = P(X \in \{51,52,\ldots,89,90\})\)

2. Problem 12

2.1. Question

Segment \(DG\) of the length \(1\) is divided at random at two points. Probability that the lengths of three subsequent (counting from point D to G) segments form a decreasing sequence, is:

  1. \(\frac{1}{2}\)
  2. \(\frac{2}{3}\)
  3. \(\frac{1}{4}\)
  4. \(\frac{1}{3}\)
  5. \(\frac{1}{6}\)

2.2. Answer

5 is true; the rest is false

3. Problem 13

3.1. Question

We toss a symmetric coin. What's the minimal number of tosses assuring with probability not less than \(0.88\) that the absolute value of difference between the frequency of heads and \(\frac{1}{2}\) is less than \(0.04\)?

  1. 216
  2. 126
  3. 378
  4. 252
  5. 567

3.2. Answer

3 is correct; the rest is wrong

Derivation:

\(X_i \in \{0,1\}\) - result of i-th cion toss; \(\mu = \frac{1}{2}\); \(\sigma^2 = \frac{1}{4}\)

\(P(|\frac{1}{n} \sum_{i=1}^n (X_i - \frac{1}{2})| < q) \geq p\)

\(|\frac{1}{n} \sum_{i=1}^n (X_i - \frac{1}{2})| = \frac{\sigma}{\sqrt{n}} |\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^n (X_i - \mu)|\)

Therefore we get:

\(P( |\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^n (X_i - \mu)| < \frac{\sqrt{n}}{\sigma} q) \geq p\)

From the **Central limit theorem** we know that our distribution can be approximated by \(N(1,0)\). Therefore:

\(2 \Phi ( \frac{\sqrt{n}}{\sigma} q) \geq p\)

And the general formula is:

\(2 \Phi ( 2 q \sqrt{n}) = p\)

For our data \(p = 0.88\), \(q = 0.04\) we get \(n = \lceil 377.7 \rceil = 378\)

4. Problem 14

4.1. Question

There are 5 ordinary dies and 2 erroneous having 6 on all sides. A randomly chosen die was rolled 4 times, and all 6 were observed. What's the probability that the die was ordinary?

  1. \(\frac{6}{2597}\)
  2. \(\frac{5}{2597}\)
  3. \(\frac{1}{2597}\)
  4. \(\frac{2}{2597}\)
  5. \(\frac{4}{2597}\)

4.2. Answer

2 is correct; the rest is wrong

It can be solved using Bayes' theorem.

5. Problem 15

5.1. Question

We calculate the integral \(I = \int_0^1 \sqrt{(sin \pi x)} dx\) using Monte-Carlo method. To do this we generate sequence \(\{X_i\}_{i \geq 1}\) of independent random variables uniformly distributed on (0;1). This integral is approximated by the quantity \(A(n) = \frac{1}{n} \sum_{i=1}^n \sqrt{sin (\pi X_i)}\). Let us notice that \(Var(\sqrt{sin(\pi X)}) \leq \frac{1}{\pi} - \frac{1}{\pi^2} = 0.21699\), \(\sqrt{0.217}=0.4658.\). 8100 simulations were performed. Absolute error \(e\) of approximation of \(I\) by \(A(n)\) is with probability \(0.94\):

  1. \(e \geq 1.9468 \times 10^{-2}\)
  2. \(e \geq 9.7342 \times 10 ^ {-3}\)
  3. \(e < 6.4894 \times 10^{-3}\)
  4. \(e = 8.0468 \times 10^{-3}\)
  5. \(e > 1.1681 \times 10^{-2}\)

5.2. Answer

2 is true; the rest is false; 1 is impossible

Problem can be solved using the Central limit theorem.

6. Problem 16

6.1. Question

Random variables \((X,Y)\) have uniform distribution on set \(D=(x,y): y \in [0,1] \wedge x \in [y-1,y]\). Error \(e\) of approximation of \(X\) by the best (in the mean square sense) Borel function \(Y\) is:

  1. \(e \in (\frac{1}{8},\frac{1}{3}]\)
  2. \(e < \frac{1}{9}\)
  3. \(e \in [\frac{1}{12},\frac{1}{8}]\)
  4. \(e > \frac{1}{5}\)

6.2. Answer

3 is true; the rest is false

\(e = V(X)(1 - \rho_{X,Y}^2)\)

\(E(X) = 0\), \(E(Y)=\frac{1}{2}\), \(E(X^2) = \frac{1}{6}\), \(E(Y^2)=\frac{1}{3}\), \(E(XY) = \frac{1}{12}\)

\(e = \frac{1}{12}\)

7. Problem 17

7.1. Question

Random variable \((X,Y)\) has uniform distribution on set \(D = \{(x,y): |x| \leq 6; 0 \leq y \leq 6-|x|\}\). Knowing that \(V(X)=6\), \(V(Y) = 2\), \(E(X^4) = \frac{432}{5}\), \(E(X^2 Y) = \frac{36}{5}\) find nonlinear regression of Y on X.

  1. \(|X|/2\)
  2. \(0\)
  3. \(\frac{36}{3}-\frac{X^2}{3}\)
  4. \(6-X\)
  5. \(\frac{6}{2} - \frac{|X|}{2}\)

7.2. Answer

5 is true; the rest is false (probably)

It can be solved by looking at the plot.

8. Problem 18

8.1. Question

Let characteristic X have distribution with the density:

\(f_p(x) = \begin{cases} \frac{p}{2|x|^{p+1}} & \text{for $|x| \geq 1$} \\ 0 & \text{for $|x| < 1$} \end{cases}\)

where \(p>3\).

It is known that: \(\int_1^\infty x f_p(x)dx = \frac{p}{2(p-1)}\), \(\int_1^\infty x^2 f_p(x)dx = \frac{p}{2(p-2)}\), \(\int_1^\infty ln(x) f_p(x) dx = \frac{1}{2p}\).

Let us denote: \(\overline{|X|}_n=\frac{1}{n} \sum_{i=1}^n |X_i|\), \(\overline{ln|X|}_n = \frac{1}{n} \sum_{i=1}^n ln|X_i|\), \(\overline{X^2}_n = \frac{1}{n} \sum_{i=1}^n X_i^2\).

Statistic \(1+\frac{1}{\overline{X}_n-1}\)

  1. is an unbiased estimator of the parameter \(p/2\)
  2. is an estimator of \(p\) having minimal variance
  3. is a consistent estimator of the parameter \(p\)
  4. is not a consistent estimator of the parameter \(p\)
  5. is an unbiased estimator of the parameter \(p\)

8.2. Answer

3 is true; the rest is false

9. Problem 19

9.1. Question

On the probability space \((<-3,3>, B(<-3,3>),\frac{1}{6}|.|)\) ( \(|.|\) - Lebesgue measure) we define a random variable:

\(X(\omega) = \begin{cases} 1 & \text{for $\omega \in [-3;-\frac{3}{2})$} \\ -2\omega & \text{for $\omega \in [-\frac{3}{2};\frac{3}{2}$} \\ -2+2\omega & \text{for $\omega \in [\frac{3}{2};3]$} \end{cases}\)

  1. mixed
  2. uniform
  3. of a continuous type
  4. of discrete type
  5. singular

9.2. Answer

1 is true; the rest is false; 2 is impossible (is it?)

10. Problem 20

10.1. Question

Let \(\{X_i\}, i \geq 1\) be a sequence of uncorrelated random variables, such that \(E(X_i)=4^{-i}; i=1,2,\cdots\), \(Var(X_i) = 2 - 5^{-i}\). \(\frac{1}{n} \sum_{i=1}^n X_i\) converges in probability to:

  1. does not converge
  2. 2
  3. 0
  4. 1
  5. -1

10.2. Answer

3 is true; the rest is false

\(\frac{1}{n} \sum_{i=1}^n X_i \rightarrow 0\)

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