Question

An unknown distribution has a mean of $100$, a standard deviation of $100$, and a sample size of $100$. Let $X=$one object of interest.

What is $P(\mathrm{\Sigma }x>9,000)$?

Short answer

The value of$\mathrm{P}(X>9000)=0.8413$.

The graph is

Step-by-step solution

Given Information

The mean is $100$, Standard deviation is $100$, and a sample size of 100.

And we need to find $P(\mathrm{\Sigma }x>9,000)$localid="1648807337983" $P(\Sigma x>9,000)$.

Calculation

The information is,

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

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

Now, we can find the probability that the sum of the $100$values is greater than $9000$

$\mathrm{P}(X>9000)=\mathrm{P}\left(Z>\frac{X-n{\mu }_x}{\sqrt{n}{\sigma }_x}\right)$

$=\mathrm{P}\left(Z>\frac{9000-10000}{1000}\right)$

$=\mathrm{P}(Z>-1)$

$\mathrm{P}(X>9000)=0.8413$

The graph of the answer is