Q. 6.99 - Elementary Statistics

Question

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.

Step-by-Step Solution

Answer

Part (a): The percentage of finishers who have times between $50$ and $70$mins is $73.01\%$.

Part (b): The percentage of finishers who have times less than $75$ mins is $94.06\%$.

Part (c): The $40$th percentile for finishers is $58.75mins$. $40\%$of finishers have time below $58.75mins$.

Part (d): The $8$th decile for finishers is $68.56 mins$. $80\%$of finishers have time below $68.56mins$.

Part (a) Step 1. Given information.

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

Part (a) Step 2. Draw the figure when time is 50 , 70 m i n s .

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 $50,70$,

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

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 $50min,70min$.

Part (b) Step 1. Draw the figure when time is 75 m i n s .

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 $94.06\%$.

Part (c) Step 1. Determine the 40 th percentile for the finishing times.

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 $58.75mins$.

Part (d) Step 1. Determine the 8 th decile for the finishing times.

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 $68.56mins$.