Ee351K Homework #9 Solution
EE 351K Probability, Statistics, and Random Processes Instructor: S.Vishwanath Homework 9 Solution
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
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