91Ó°ÊÓ

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

Formula used:$\text{population mean and standard deviation:}{\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}\text{and}{\mathrm{σ}}_{\hat{x}}=\mathrm{σ}/\sqrt{n}\text{.}$

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

${\mathrm{μ}}_{\stackrel{¯}{\stackrel{¯}{x}}}=\mathrm{μ}\&{\mathrm{σ}}_{\stackrel{¯}{\stackrel{¯}{x}}}=\frac{\mathrm{σ}}{\sqrt{n}},n=$sample size

$∴$Mean of $\stackrel{¯}{x}={\mathrm{μ}}_{\stackrel{¯}{x}}$

$=\mathrm{μ}$

$=61\min$

$\&S.D$of $\stackrel{¯}{x}={\mathrm{σ}}_{\stackrel{¯}{x}}$

$=\frac{\mathrm{σ}}{\sqrt{4}}$

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

$=4.5\min$

$∴$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 $≤$size $n=49$

$∴$${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$

$$\begin{gathered} =61min \\ {\mathrm{σ}}_{\stackrel{¯}{\stackrel{¯}{X}}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{9}{\sqrt{9}} \\ =3min \end{gathered}$$

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, $P[\mathrm{μ}-5≤\stackrel{¯}{X}≤\mathrm{μ}+5]$

Where $\stackrel{¯}{X}=$sample mean

$n=$sample size

$=4$

${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}=61\min$

${\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}=4.5\min$

$P[\mathrm{μ}-5≤\stackrel{¯}{X}≤\mathrm{μ}+5]$

$=P\left[\frac{\mathrm{μ}-5-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\mathrm{μ}+5-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}\right]$[Subtracting $\mathrm{μ}$from every term in

The inequality $\&$then dividing

By ${\mathrm{σ}}_{\stackrel{¯}{x}}$

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

Where $$\begin{gathered} z=\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}\&z~N(0,1) \\ âˆ\mu \stackrel{¯}{X}~N\left(\mathrm{μ},{\mathrm{σ}}_{\stackrel{¯}{X}}\right) \\ =P[-1.11≤z≤1.11] \\ =P[z≤1.11]-P[z≤-1.11] \\ =\mathrm{Φ}(1.11)-\mathrm{Φ}(-1.11) \\ =2\mathrm{Φ}(1.11)-1 \\ =2×(0.8665005)-1 \\ =0.733001 \\ =73.3001\% \end{gathered}$$

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, $P[\mathrm{μ}-5≤\stackrel{¯}{X}≤\mathrm{μ}+5]$

Where ${\mathrm{μ}}_{\stackrel{¯}{X}}=\mathrm{μ}=61\min \&{\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}=\frac{9}{\sqrt{9}}=3\min$

$$\begin{gathered} P[\mathrm{μ}-5≤\stackrel{¯}{X}<\mathrm{μ}+5] \\ =P\left[-\frac{5}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{5}{{\mathrm{σ}}_{\stackrel{¯}{X}}}\right] \\ =P\left[-\frac{5}{3}≤z≤\frac{5}{3}\right],z=\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}~N(0,1) \\ =P[-1.67≤z≤1.67] \\ =\mathrm{Φ}(1.67)-\mathrm{Φ}(-1.67) \\ =2×\mathrm{Φ}(1.67)-1 \\ =2×9525403-1 \\ =.9051 \\ =90.51\% \end{gathered}$$

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