Exercise 7.17 THE BANK CUSTOMER WAITING TIME CASE Recall that the bank manager wants to show that the new system reduces typical customer waiting times to less than six minutes. One way to do this is to demonstrate that the mean of the population of all customer waiting times is less than 6.1. Letting this mean be µ, in this exercise we wish to investigate whether the sample of 101 waiting times provides evidence to support the claim that µ is less than 6.1. | For the sake of argument, we will begin by assuming that µ equals 6.1, and we will then attempt to use the sample to contradict this assumption in favor of the conclusion that µ is less than 6.1. Recall that the mean of the sample of 101 waiting times is = 5.5 and assume that σ, the standard deviation of the population of all customer waiting times, is known to be 2.47 | (a) | Consider the population of all possible sample means obtained from random samples of 101 waiting times. What is the shape of this population of sample means? That is, what is the shape of the sampling distribution of ? | Normal because the sample is large. | (b) | Find the mean and standard deviation of the population of all possible sample means when we assume that µ equals 6.1. (Round your answer to 3 decimal places.) | µ = 6.1, σ = |

(c) | The sample mean that we have actually observed is = 5.5. Assuming that µequals 6.1, find the probability of observing a sample mean that is less than or equal to = 5.5. (Round your answer to 4 decimal places.) | P( < 5.5) | | (d) | If µ equals 6.1, what percentage of all possible sample means are less than or equal to 5.5? What do you conclude about whether the new system has reduced the typical customer waiting time to less than six minutes? (Round your answer to 2 decimal places.) | %; conclude that µ is less than 6.1. |
USING THE EXCEL TEMPLATE
Use the tab for Sample Mean Distribution

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| | | | Distribution of Sample Mean | : | | | | | | | | | | | | | | | | | | | Popn. Mean | 6.1 | μ | | | | | | Popn. Stdev. | 2.47 | σ | | | | | | Sample Size | 101 | n | | | | | | | | | | | | | | Confidence Intervals, 2 tailed | | | | | Distn. of X-bar | | | a | x1 | x2 | | | | | | Mean | Stdev | | | | | | | | | | | 6.1 | 0.2458 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Inverse calculation | | | x1 | x2 | P(x1<Xbar<x2) | | x | P(<x) | P(>x) | P(<x) | x | P(>x) | x | | | | | | 5.5 | 0.0073 | | | | | | | | | | | | | | | | | |
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Exercise 6.45 METHODS AND APPLICATIONS

Suppose that x has a binomial distribution with n = 200 and p = 0.3. (Round your answers to 4 decimal places. Round z values to 2 decimal places.) | (a) | Show that the normal approximation to the binomial can appropriately be used to calculate probabilities about x. | | | | np | = | | n(1 – p) | = | | |

Both np and n(1 – p) > 5 | (b) | Make continuity corrections for each of the following, and then use the normal approximation to the binomial to find each probability: | | | | | 1. | P (x = 60) | = | | 2. | P (x ≤ 70) | = | | 3. | P (x < 50) | = | | 4. | P (x ≥ 75) | = | | 5. | P (x > 75) | = | |