# A population of values has a normal distribution with \mu=116.3 and \sigma=27.5. You intend to draw a random sample of size n=249. Find the probabilit

A population of values has a normal distribution with $\mu =116.3$ and $\sigma =27.5$. You intend to draw a random sample of size $n=249$.
Find the probability that a sample of size $n=249$ is a randomly selected with a mean greater than 117.3.
$P\left(\stackrel{―}{X}>117.3\right)=$?
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Aamina Herring
$\stackrel{―}{X}=\frac{1}{n}\sum _{i=1}^{n}{X}_{i}=$
Mean value.
Therefore, the mean of $\stackrel{―}{X}$ is 116.3,
standard deviation is $\frac{27.5}{\sqrt{n}}=\frac{27.5}{\sqrt{249}}$, as $n=249$.
The probability of a mean greater than 117.3 is
$P\left[\stackrel{―}{X}>117.3\right]$
$=P\left[\frac{\stackrel{―}{X}-116.3}{\frac{27.5}{\sqrt{249}}}>\frac{117.3-116.3}{\frac{27.5}{\sqrt{249}}}\right]$
$=P\left[Z>\frac{\sqrt{249}}{27.5}\right],Z=\frac{\stackrel{―}{X}-116.3}{\frac{27.5}{\sqrt{249}}}\sim$ Normal(0,1)
$=1-P\left[Z\le \frac{\sqrt{249}}{27.5}\right]$
$=1-\varphi \left(0.574\right)$
$=1-0.717=0.283$
The value of $\varphi \left(0.574\right)$ is taken from the Normal distribution table.
Therefore, the probability of the mean is greater than 117.3 is 0.283.