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Q.11

Expert-verifiedFound in: Page 428

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

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

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

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

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

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

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

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