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Q.11

Expert-verified
Found in: Page 428

### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208

# 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 $\mathrm{P}\left(X\ge 7500\right)=0.0089$.

See the step by step solution

## Step 1: Given Information

A mean (${\mu }_{x}$) is$180$ and a standard deviation (${\sigma }_{x}$) is $20$ and sample size $\left(n\right)$ is$40$.

## Step 2: Explanation

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

$\sum X~N\left(n{\mu }_{x},\sqrt{n}{\sigma }_{x}\right)$

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

$\mathrm{P}\left(X\ge 7500\right)=\mathrm{P}\left(Z\ge \frac{X-n{\mu }_{x}}{\sqrt{n}{\sigma }_{x}}\right)=\mathrm{P}\left(Z\ge \frac{7500-7200}{126.4911}\right)=\mathrm{P}\left(Z\ge 2.3717\right)$

$\mathrm{P}\left(X\ge 7500\right)=0.0089$