New York City $10-km$Run. As reported by Rumле's World magazine, the times of the finishers in the New York City 10-km run are normally distributed with a mean of 61 minutes and a standard deviation of $9$minutes. Do the following for the variable "finishing time$61min$ of finishers in the New York City $10-\mathrm{km}$run.

a. Find the sampling distribution of the sample mean for samples of size $4$

b. Repeat part (a) for samples of size $9$

C. Construct graphs similar to those shown in Fig.$7.4$on-page $304$

d. Obtain the percentage of all samples of four finishers that have mean finishing times within $5$minutes of the population mean finishing time of 61 minutes. Interpret your answer in terms of sampling error.

e. Repeat part (d) for samples of size $9$

$\text{Times of the finishers are normally distributed with mean}\mu =61\min \&\mathrm{S}.\mathrm{D}\sigma =9\min$

Formula used:$\text{population mean and standard deviation:}{\mu }_{\overline{x}}=\mu \text{and}{\sigma }_{\hat{x}}=\sigma /\sqrt{n}\text{.}$

The sample mean $\overline{x}$of size $4$samples will follow a normal distribution with mean

${\mu }_{\overline{\overline{x}}}=\mu \&{\sigma }_{\overline{\overline{x}}}=\frac{\sigma }{\sqrt{n}},n=$sample size

$\therefore$Mean of $\overline{x}={\mu }_{\overline{x}}$

$=\mu$

$=61\min$

$\&S.D$of $\overline{x}={\sigma }_{\overline{x}}$

$=\frac{\sigma }{\sqrt{4}}$

$=\frac{9}{2}\min$

$=4.5\min$

$\therefore$The mean finishing time for a sample of size $4$follows a normal distribution, with a mean of $61$ and a standard deviation of $61\&S.D4.5min$

Here sample $\le$size $n=49$

$\therefore$${\mu }_{\overline{x}}=\mu$

$$\begin{gathered} =61min \\ {\sigma }_{\overline{\overline{X}}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{9}{\sqrt{9}} \\ =3min \end{gathered}$$

As a result, the sample mean follows the typical distance, with a mean of $61\min \&s.d3min$

The figure is

We have to find, $P[\mu -5\le \overline{X}\le \mu +5]$

Where $\overline{X}=$sample mean

$n=$sample size

$=4$

${\mu }_{\overline{x}}=\mu =61\min$

${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}=4.5\min$

$P[\mu -5\le \overline{X}\le \mu +5]$

$=P\left[\frac{\mu -5-\mu }{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\le \frac{\mu +5-\mu }{{\sigma }_{\overline{X}}}\right]$[Subtracting $\mu$from every term in

The inequality $\&$then dividing

By ${\sigma }_{\overline{x}}$

$=P\left[-\frac{5}{4.5}\le z\le \frac{5}{4.5}\right]$

Where $$\begin{gathered} z=\frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\&z~N(0,1) \\ \because \overline{X}~N\left(\mu ,{\sigma }_{\overline{X}}\right) \\ =P[-1.11\le z\le 1.11] \\ =P[z\le 1.11]-P[z\le -1.11] \\ =\Phi (1.11)-\Phi (-1.11) \\ =2\Phi (1.11)-1 \\ =2\times (0.8665005)-1 \\ =0.733001 \\ =73.3001\% \end{gathered}$$

As a result, $73.3001\%$ of all feasible size $4$ samples will finish within $5$ minutes of the population mean finishing time of $61\min$

Here the sample size $n=9$

We have to find, $P[\mu -5\le \overline{X}\le \mu +5]$

Where ${\mu }_{\overline{X}}=\mu =61\min \&{\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}=\frac{9}{\sqrt{9}}=3\min$

$$\begin{gathered} P[\mu -5\le \overline{X}<\mu +5] \\ =P\left[-\frac{5}{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\le \frac{5}{{\sigma }_{\overline{X}}}\right] \\ =P\left[-\frac{5}{3}\le z\le \frac{5}{3}\right],z=\frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}~N(0,1) \\ =P[-1.67\le z\le 1.67] \\ =\Phi (1.67)-\Phi (-1.67) \\ =2\times \Phi (1.67)-1 \\ =2\times 9525403-1 \\ =.9051 \\ =90.51\% \end{gathered}$$

As a result, $9051\%$of all conceivable samples of size $9$will complete within $5$minutes of the population's average finish time of$61min$