Please help asap and detailed Calculation problems 1. Suppose that first examine

Question

Please help asap and detailed Calculation problems 1. Suppose that first examinees pass an exam with probability 0.6. Whenever they succeed, they do not have another go. If an exam was unsccssu, they attempt it at the next occasion. However, because they study a bit more, the chance of them to fail is half the probability at their previous try. Everyone can have at most three attempts. Let X denote the number of tests taken by a random student (a) What is the probability that someone fulfils the requirements? (2 pts) (b) Give the distribution of X (5 pts) (c) Calculate the mean value and the standard deviation of X (5 pts) (d) 225 students are to pass the exam. What is the mean value and standard deviation of the number of necessary test sheets? Give the exact probability that more than 350 sheets will be needed, and calculate (6 pts) its approximate value.

Explanation / Answer

Question 1

Here the probability that someone fills thr requirements = Pr(SOmeone pass the exam) = Pr(pass in first attempt) + Pr(Pass in second attempt) + Pr(Pass in third attempt)

= 0.6 + 0.4 * (1 - 0.4/2) + 0.4 * (0.4/2) * (1 - 0.4/2 + 0.4/4) = 0.992

(b) Here

p(X) = 0.6 ; x = 1

= 0.4 * (1 - 0.4/2) =0.32 ; x = 2

= 1 - 0.6 - 0.32 = 0.08 ; x = 3

(c) E[X] = 1 * 0.6 + 2 * 0.32 + 0.08 * 3 = 1.48

Var[X] = E[X2] - E[X]2

E[X2] = 12 * 0.6 + 22 * 0.32 + 32 * 0.08 = 2.6

Var[X] = 2.6 - 1.482 = 0.4096

SD[X] = sqrt(0.4096) = 0.64

(d) Her if 225 students are to pass the exam.

Here the mean value of test sheets needed = 225 * 1.48 = 333

Standard deviation of test sheets needed = sqrt(225) * 0.64 = 9.6

Pr(X > 350 ; 333 ; 9.6) = 1 - Pr(X <= 350 ; 333 ; 9.6)

Z = (350 - 333)/9.6 = 1.77083

Pr(X > 350 ; 333 ; 9.6) = 1 - Pr(X <= 350 ; 333 ; 9.6) = 1 - Pr(Z < 1.77083) = 1 - 0.9617 = 0.0383

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