Question

Find the sum that is one standard deviation above the mean of the sums.

Short answer

The required sum is$\sum X=7326.49$.

Step-by-step solution

Given Information

A sample size $\left(n\right)=40$.

The mean of population $\left({\mu }_X\right)=180$.

$z=1$.

A standard deviation $\left({\sigma }_X\right)=20$.

Explanation

To find the sum that is one standard deviation above the mean of the sums :

$\sum X=\left(n\right)\left({\mu }_X\right)+\left(z\right)\left(\sqrt{n}\right)\left({\sigma }_X\right)$

Calculating, we have:

$\sum X=40\times 180+1\times \sqrt{40}\times 20$

$\sum X=7326.49$