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Q.11

Expert-verified

Introductory Statistics

Found in: Page 428

Book edition OER 2018

Author(s) Barbara Illowsky, Susan Dean

Pages 902 pages

ISBN 9781938168208

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Short Answer

Find the probability that the sum of the $40$ values is greater than $7, 500$ .

The probability that the sum of the $40$ values is greater than $7500$ is $P (X \geq 7500) = 0.0089$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

A mean ( $μ_{x}$ ) is $180$ and a standard deviation ( $σ_{x}$ ) is $20$ and sample size $(n)$ is $40$ .

Step 2: Explanation

To find the probability that the sum of the $40$ values is less than that $7500$ :

$\sum X ~ N (n μ_{x}, \sqrt{n} σ_{x})$

$\sum X ~ N ((40) (180), (\sqrt{40}) (20))$

$P (X \geq 7500) = P (Z \geq \frac{X - n μ_{x}}{\sqrt{n} σ_{x}}) = P (Z \geq \frac{7500 - 7200}{126.4911}) = P (Z \geq 2.3717)$

$P (X \geq 7500) = 0.0089$

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