Question

Find the probability that the sum of the $100$values is less than 3900.

Short answer

The probability that the sum of the $100$ values is less than $3,900$isrole="math" localid="1648552619706" $\mathrm{P}(X\le 3900)=0.4207$.

Step-by-step solution

Given Information

The mean $\left({\mu }_x\right)$is $39.01$with a standard deviation ${\sigma }_x$is $0.5$. The researcher randomly selects a sample$\left(n\right)$ is$100$.

Explanation

To find the probability that the sum of the $100$values is greater than $3900$:

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

$\sum X~N\left(\right(100\left)\right(39.01),(\sqrt{100}\left)\right(0.5\left)\right)$

localid="1648552431433" $\mathrm{P}(X\le 3900)=\mathrm{P}\left(Z\le \frac{X-n{\mu }_x}{\sqrt{n}{\sigma }_x}\right)=\mathrm{P}\left(Z\le \frac{3900-3901}{5}\right)=\mathrm{P}(Z\le -0.2)$$\mathrm{P}(X\le 3900)=0.4207$