Question

robability Theory and Mathematical statistics Final examination Variant 4 Part 1. Random Events he probability that a computer crashes during a severe thunderstorm is 0.005. A certain npany had 550 working computers when the area was hit by a severe thunderstorm. Compute ne probability that exactly 2 computers crashed. 2. It is known about random events A and B that PCB) = 5P (AB). PCA) = 0.7and P(A + B) = 0.6. Find P(B). 3. A problem on a multiple-choice quiz is answered correctly with probability 0.75 if a student is prepared. An unprepared student guesses between 5 possible answers, so the probability of choosing the right answer is 1/5. 80% of students are prepared for the quiz. If Mr.X gives a wrong answer to the problem, what is the probability that he is prepared for the quiz? Part 2. Random Variables 4. Two independent random variables X and Y are given with their distribution laws X -8 4 Yi -2 0 2 Pi 0.6 0.4 Pi 0.1 0.2 0.7 Find the distribution law of the random variable W=X2 +Y 5. Let X be a continuous random variable with the Normal distribution NC2; 9). Find the probability that the variable of X does not belong to interval (2, 4). 6. The cumulative distribution function of random variable Xis! 0, if ISO F(x) = a + b sinzx, if 0<x<1/12 Fin coefficients a and b. 1 7. Let X be a continuous random variable with probability density function: f(x) = (x2 - 1), if *<2 lo, if x 22 Find the variance of X Part 3. Statistics The following sample is given 1 5 6 5 6 4 6 7 8. Find the empirical distribution law and draw frequency polygon 9. Find 90% confidence interval for the population variance.

6th

pls answer it fast

Answer 1

Since, F(x) is a distribution function, it is right continuous, that is F(x+)=F(x) for every x. For the given F(x), F(O)=0 and F(0+)=lim F(e) = lim(a + bSin(2e)) = a Since, F is right continuous, we have F(O+)=F(O), that is, a=0. Again, F is the DF of a continuous random variable, so that it is left continuous also. That is, for every x, F(x)-F(x-)=0. Now for the given F(x), FC=lim F (. –e) = lim ( a + bsin (268 – c)))

= 0 + b lim Sin (7 – 2e) = bSin () = b/2. Since, F (72) - F 65,-) = 0 we get 1 – = 0 or b = 2. Thus a=0 and b=2.

The answer is based on the facts that i) All distribution functions are right continuous and (ii) the distribution function of a continuous random variable is left continuous also.

For any query in above, comment.