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Chapter 7: Q. 7.9 (page 414)

Use the information in Example \(7.9\), but change the sample size to \(144\).

a. Find \(P(20<\bar{x}<30)\).

b. Find \(P(\sum x\) is at least \(3,000)\).

c. Find the \(75th\) percentile for the sample mean excess time of \(144\) customers.

Short Answer

Expert verified

Part a. \(P(20<\bar{x}<30)\)

Part b. \(P(\sum x\geq 3000)=0.7377\)

Part c. \(75th\)percentile\(=23.2\)

Step by step solution

01

Part a. Step 1. Given information

Mean excess time used \(=22\)minutes

Sample size \(=144\)

02

Part a. Step 2. Calculation

\(\bar{X}=N(22,\frac{22}{\sqrt{144}})\)

\(P(20<\bar{x}<30)=normalcdf(20,30,22,\frac{22}{\sqrt{144}})\)

\(P(20<\bar{x}<30)=0.8623\)

03

Part b. Step 1. Calculation

\(\sum X=N((144)(22),(\sqrt{144})(22))\)

\(P(\sum x\geq 3000)=normalcdf(3000,E99,(144)(22),(\sqrt{144})(22))\)

\(P(\sum x\geq 3000)=0.7377\)

04

Part c. Step 1. Calculation

\(75th\)percentile\(=invNorm(\frac{75}{100},22,\frac{22}{\sqrt{144}})\)

\(75th\)percentile\(=23.2\)

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