# Ee351K Homework #9 Solution

**Topics:**Probability theory, Random variable, Variance

**Pages:**5 (1051 words)

**Published:**December 4, 2012

FALL 2012 sriram@ece.utexas.edu

Problem 1 A fair coin is tossed 100 times. Estimate the probability that the number of heads lies between 40 and 60 using central limit theorem(the word between in mathematics means inclusive of the endpoints). Solution: The expected number of heads is 100 1 = 50, and the variance for the number of heads is 2 11 100 2 2 = 25. Thus, since n = 100 is reasonably large, we have Sn − 50 ∼ N (0, 1) 5 40 − 50 60 − 50 ∗ P (40 ≤ Sn ≤ 60) = P ≤ Sn ≤ 5 5 ∗ = P (−2 ≤ Sn ≤ 2) ∗ Sn =

= 2[φ(2) − 0.5] = 2 ∗ (.4772) = .9544 Problem 2 An insurance company has 10,000 automobile policyholders. The expected yearly claim per policyholder is $240 with a standard deviation of $800. Approximate the probability that the total yearly claim exceeds $2.7 million. Solution: Let Cj denote the annual claim made by the j th policyholder in dollars. We are given that E[Cj ] = 240 and V ar(Cj ) = (800)2 for each j = 1, 2, ..., 10000, (since the annual claims made by the policyholder are identical copies of each other). Let T = Cj , which represents the total claim amount from all 10,000 policyholders. We want to ﬁnd P (T > 2, 700, 000) . Since the sample size is sufﬁciently large the Central Limit Theorem tells us that T ∼ N (2, 400, 000, 10, 000(800)) . Using this fact we may approximate the probability to get the following: P (T > 2700000) = P T − 2400000 2700000 − 2400000 > 800000 800000 = P (Z > .375) = .3538

Problem 3 A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and its standard deviation is 30 hours, how many of these components must be in stock so that the probability that the system is in continual operation for the next 2,000 hours is at least .95? Solution: Let Lj denote the lifetime of the jth component used in the electrical system. We are given that E[Lj ] = 100hrs. and V ar(Lj ) = 900hrs2 . for each j = 1,...,n (since the components are identical copies of each other). Let Rn = Lj , which denotes the total lifetime of the system when n components are available for use. We need to ﬁnd n such that P (Rn > 2000hr.s) ≥ .95 . Since we don’t know the distribution of the individual component lifetimes, only method for ﬁnding the critical n is to use a Central Limit Theorem approximation. By the CLT, Rn ∼ N (100nhrs., 900nhrs.). Using this information we can set up the following calculation.

.95 ≤ P (Rn > 2000hr.s) = P P Z>

Rn − 100n 2000 − 100n √ √ > 30 n 30 n

= P (Z >

Rn − 100n √ ) 30 n

2000 − 100n √ ≥ 0.95 30 n 2000 − 100n √ ≤ −1.645 30 n Rn ≥ 22.3

Rounding up we now have that the number of components available for use must be at least 23. Problem 4 From past experience a professor knows that the test score of a student taking her ﬁnal examination is a random variable with a mean of 75 and standard deviation of 8. How many students would have to take the examination to ensure, with probability at least .95 that the class average would be at least 73? Problem 5 Let X denote the average test score based on n students. From the CLT we know that if the number of students is sufﬁciently large X ∼ N (75, 64 ) . We need to ﬁnd n such that P (X > 73) > .95. n Since we don’t know the exact distribution of X we must use the CLT approximation to calculate the probability and ﬁnd the critical value of n. √ − n ) = P (Z ≥ 4

0.95 ≤ P (X > 73) = P

X − 75 73 − 75 √ > √ 8/ n 8/ n

Looking at the above line for a moment we notice that this is true only (Looking in the Normal table) when √ − n 4 < −1.645. We are then left with an inequality involving only n from which it follows:n > 43.29, n = 44 Rounding up we see that if at least 44 students take the exam the conditions will be satisﬁed. Problem 5 Let Xi , i = 1, 2, . . . be i.i.d. continuous random variables that...

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