# A population of values has a normal distribution with mean 191.4 and standard deviation of 69.7. A random sample of size n = 153 is drawn. Find the probability that a sample of size n=153 is randomly selected with a mean between 188 and 206.6. Round your answer to four decimal places. P=?

A population of values has a normal distribution with mean 191.4 and standard deviation of 69.7. A random sample of size $n=153$ is drawn.
Find the probability that a sample of size $n=153$ is randomly selected with a mean between 188 and 206.6. Round your answer to four decimal places.
P=?
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The probability that a sample size $n=153$ is randomly selected with a mean between 188 and 206.6 is obtained below:
From the given information, a population of values has a normal distribution with mean 191.4 and standard deviation of 69.7 and sample of size $n=153$.
here,
if $X\sim \left(\mu ,\sigma \right)$ then $\stackrel{―}{X}\sim \left(\mu ,\frac{\sigma }{\sqrt{n}}\right)$
The required value is given below:
$P\left(188\le \stackrel{―}{X}\le 206.6\right)=P\left(\stackrel{―}{X}\le 206.6\right)-P\left(\stackrel{―}{X}\le 188\right)$
$=P\left(Z\le \frac{206.6-191.4}{\frac{69.7}{\sqrt{153}}}\right)-P\left(Z\le \frac{188-191.4}{\frac{69.7}{\sqrt{153}}}\right)$
$=P\left(Z\le \frac{15.2}{5.635}\right)-P\left(Z\le \frac{-3.4}{5.635}\right)$
$=P\left(Z\le 2.70\right)-P\left(Z\le -0.60\right)$
$=0.9965-0.2743=0.7222$
Thus, the probability that a sample size $n=153$ is randomly selected with a mean between 188 and 206.6 is 0.7222.
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