As reported in Runner's World magazine, the times of the finishers in the New York City $10$-km run are normally distributed with mean $61$minutes and standard deviation 9 minutes.

Part (a): Determine the percentage of finishers who have times between $50$and $70$minutes.

Part (b): Determine the percentage of finishers who have times less than $75$minutes.

Part (c): Obtain and interpret the $40$th percentile for the finishing times.

Part (d): Find and interpret the $8$th decile for the finishing times.

The given mean is $61min$ and standard deviation is$9min$.

Draw the figure showing the required shaded region and its delimiting x-values, which are $50,70$.

We need to compute the z-scores for the x-values $50,70$,

localid="1652474127361" $$\begin{gathered} x=50\to \\ z=\frac{50-61}{9} \\ =-1.22 \\ x=70\to \\ z=\frac{70-61}{9} \\ =1 \end{gathered}$$

We need to find the area under the standard normal curve that lies between $-1.22,1$. The area to the left of $-1.22$is $0.1112$and the area to the left of $1$is $0.8413$. The required area shaded is between $0.8413-0.1112=0.7301$.

On interpreting, we can say, $73.01\%$of finishers have times between $50min,70min$.

Draw the figure showing the required shaded region and its delimiting x-values, which are $75mins$.

We need to compute the z-scores for the x-values $75$,

$$\begin{gathered} x=75\to \\ z=\frac{75-61}{9} \\ =1.56 \end{gathered}$$

We need to find the area under the standard normal curve that lies below $1.56$. The area to the left of $1.56$is $0.9406$. The required area shaded is $0.9406$.

The required percentage is$94.06\%$.

The z-score corresponding to $P_{40}$is the one having an area $0.4$ to its left under the standard normal curve. From standard normal table, that z-score is $0.25$.

Now, we must find the x-value having the z-score $0.25$, the length is $0.25$standard deviations below the mean.

It is $61-0.25\left(9\right)=58.75$.

The $40th$percentile for finishers is $58.75mins$.

On interpreting, we can say, $40\%$of finishers have time below$58.75mins$.

The z-score corresponding to $P_{80}$, eight decile is the one having an area $0.8$ to its left under the standard normal curve. From standard normal table, that z-score is $0.84$.

Now, we must find the x-value having the z-score $0.84$, the length is $0.84$standard deviations below the mean. It is $61+0.84\left(9\right)=68.56$.

The $8th$decile for finishers is $68.56$minutes.

On interpreting, we can say, $80\%$of finishers have time below$68.56mins$.