## Related questions with answers

Based on the data in table below, assume that ages of moviegoers are normally distributed with a mean of 35 years and a standard deviation of 20 years.

$\begin{array} { c c c c c c c c }\hline\text{Age}& { 2 - 11 } & { 12 - 17 } & { 18 - 24 } & { 25 - 39 } & { 40 - 49 } & { 50 - 59 } &\text{60 and older}\\ \hline\text{Percent}&{ 7 } & { 15 } & { 19 } & { 19 } & { 15 } & { 11 }&14 \\ \hline \end{array}$

Find the probability that a simple random sample of 25 moviegoers has a mean age that is less than 30 years.

Solution

VerifiedGiven:

$\begin{align*} \mu&=\text{Mean}=35 \\ \sigma&=\text{Standard deviation}=20 \\ n&=25 \\ \overline{x}&=30 \end{align*}$

We need to determine the probability that the mean age $\overline{X}$ of moviegoers is less than 30 years.

$P(X<30)$

The sampling distribution of the sample mean has mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$.

The z-value is the sample mean decreased by the population mean, divided by the standard deviation:

$z=\dfrac{\overline{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{30-35}{20/\sqrt{25}}=\dfrac{-5}{20/5}=\dfrac{-5}{4}=-1.25$

Let us determine the corresponding probability using the normal probability table in the appendix, which is given in the row starting with "$-0.2$" and in the column starting with ".05".

$P(\overline{X}<30)=P(Z<-1.25)=0.1056=10.56\%$

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