Question

Find the probability that the sum of the $100$values is greater than $3,910$.

Short answer

The probability that the sum of the $100$ values is greater than $3,910$is$\mathrm{P}(X\ge 3910)=0.0359$.

Step-by-step solution

Given Information

From the information given in the question, the mean is $39.01$with a standard deviation of $0.5$.

Explanation

To find the probability that the sum of the $100$values is greater than$3910$:

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

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

$\mathrm{P}(X\ge 3910)=\mathrm{P}\left(Z\ge \frac{X-n{\mu }_x}{\sqrt{n}{\sigma }_x}\right)=\mathrm{P}\left(Z\ge \frac{3910-3901}{5}\right)=\mathrm{P}(Z\ge 1.8)$

$\mathrm{P}(X\ge 3910)=0.0359$