avissidep

2020-12-22

A population of values has a normal distribution with $\mu =77$ and $\sigma =32.2$. You intend to draw a random sample of size $n=15$
Find the probability that a sample of size $n=15$ is randomly selected with a mean between 59.5 and 98.6. $P\left(59.5<\stackrel{―}{X}<98.6\right)=$?

faldduE

From the given information, $\mu =77$, $\sigma =32.2$ and the sample size=15.
Consider,
$P\left(59.5<\stackrel{―}{X}<98.6\right)=P\left(\frac{59.5-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<\frac{\stackrel{―}{X}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<\frac{98.6-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right)$
$=P\left(\frac{59.5-\mu }{\left(\frac{32.2}{\sqrt{15}}\right)}
$=P\left(-2.10488
$=P\left(z<2.59803\right)-P\left(z<-2.10488\right)$

Thus, the probability that a sample of size $n=15$ is randomly selected with a mean between 59.5 and 98.6 is 0.9777.

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