# 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 single randomly selected value is greater than 117.3. P(X > 117.3) =? Write your answers as numbers accurate to 4 decimal places.

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 single randomly selected value is greater than 117.3.
$P\left(X>117.3\right)=$?
Write your answers as numbers accurate to 4 decimal places.
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Step 1
Suppose, X follows Normal distribution with mean $\mu =116.3$ and standard deviation $\sigma =27.5$.
The z-score for x is defined as,
$Z=\frac{x-\mu }{\sigma }$
$=\frac{x-116.3}{27.5}$
Step 2
The probability of a single randomly selected value is greater than 117.3
$P\left[X>117.3\right]$
$=P\left[\frac{X-116.3}{27.5}>\frac{117.3-116.3}{27.5}\right]$
$=P\left[Z>\frac{1}{27.5}\right]$
$=1-P\left[Z\le \frac{1}{27.5}\right]$
$=1-\varphi \left(0.036\right)$
$=1-0.5144=0.4856$
The value of $\varphi \left(0.036\right)$ is taken from the Normal distribution table.
Therefore, the probability of a single randomly selected value is greater than 117.3 is 0.4855.