Question

An unknown distribution has a mean of $80$and a standard deviation of $12$. A sample size of $95$is drawn randomly from the population.

Find the sum that is two standard deviations above the mean of the sums.

Short answer

The sum that is two standard deviations above the mean of the sums is$7833.92$

Step-by-step solution

Given Information

Given in the question that,

mean is$80$, standard deviation is$12$and sample size is$95$

Find the sum that is two standard deviations above the mean of the sums.

Explanation

According to the shared details, the sample size of $95$is randomly drowned from the population with a mean of $80$and a standard deviation is $12$. Therefore, the sum that is $2$standard deviations above the mean of sums is given as:

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

$=95\times 80+2\times \sqrt{95}\times 12$

$=7833.92$