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 probability that the sum of the $95$values is less than $7400$.

Short answer

The probability that the sum of the $95$values is less than $7400$is .

$\mathrm{P}(X\le 7400)=0.0436$

Step-by-step solution

Given Information

Given in the question that,

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

Find the probability that the sum of the $95$values is less than $7400$.

Explanation

From Example, we have the next information

$\sum X~N(n{\mu }_x,\sqrt{n}{\sigma }_x)$

$\sum X~N\left(\right(95\left)\right(80),(\sqrt{95}\left)\right(12\left)\right)$

Now, we can find the probability that the sum of the $95$values is less than $7400$

$\mathrm{P}(X\le 7400)=\mathrm{P}(Z\le \frac{X-n{\mu }_x}{\sqrt{n}{\sigma }_x})=\mathrm{P}(Z\le \frac{7400-7600}{116.961532})=\mathrm{P}(Z\le -1.71)$

$\mathrm{P}(X\le 7400)=0.0436$

Explanation

The following is obtained graphically,