HW Assignment #3 Assigned: December 11th - Firm Due Date: 18/12/2017 by Midnight
ID: 3325702 • Letter: H
Question
HW Assignment #3 Assigned: December 11th - Firm Due Date: 18/12/2017 by Midnight Please clearly justify your answers Solutions containing only formulas or only final answers will get: 0 points Submit your solutions in soft copy to the Blackboard before specified deadline You should submit a clean scanned copy to the Blackboard Problem #1: (10pts) The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviati observed. Find the probability that the average time waiting in line for these customers is (a) Less than 10 minutes (b) Between 5 and 10 minutes (c) Less than 6 minutes 5 minutes. Suppose that a randoms ample of n 49 customers is Problem 2: A consumer electronics company is comparing the brightness of two different types of picture tubes for use in its television sets. Tube type A has mean brightness of 100 and standard deviation of 16, and tube type B has unknown mean brightness, but the standard deviation is assumed to be identical to that for type A. A random sample of n 25 tubes of each type is selected, and XB - JA is computed. If uB equals or exceeds , the manufacturer would like to adopt type B for use. The observed difference is xB-Xa decision would you make, and why? = 3.5. What Problem 3: Suppose that .and@2 are estimators of the parameter We know that E(A)- , E(A) = /2. V(A) = 10, and V(62)-4 Which estimator is better? In what sense is it better? (in terms of unbiasedness and relative efficiency)Explanation / Answer
Q1. (a) The mean waiting time for the customer is 8.2 min and std. dev of 1.5 min. For a random sample of 49 customers
prob that average waiting time is 10 min so we have to compute a z statistic where z = x-u/std. dev. = 10 - 8.2 / 1.5 = 1.2. So we need to find the p value which is area under the std. normal curve less that z=1.2
Using technology like matlab we find p = 0.8849 so there is 88%chance that customer has to wait less than 10 min
(b) rob that waiting time is beween 5 and 10 min is area under the curve between z= 5-8.2 / 1.5 and z=10-8.2/1.5
i.e. z = -2.13 and z =1.2 i.e. area to the left of z=1.2 - area to the left of z=-2.13 = 0.8849 - 0.016 = 0.8689
So there is 86% chance of waiting beween 5 and 10 min
(c) Prob that waiting time is less than 6 min is z = (6 - 8.2)/1.5 = -1.46
So area to the left of z = -1.46 on the std. normal curve computed using matlab = 0.072
Hence there is 7 % chance of waiting time being less than 6 min
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