Chapter PI, Problem 22RE

(a)

To determine

To find out what percent of the times should be within $0.5\text{ second of 78}\text{.21}$ .

Answer to Problem 22RE

  $37.59\%$ .

Explanation of Solution

In the question, it is given the table showing the top women’s skating times. Thus, we know that the mean finishing time was $78.21$ seconds and the standard deviation is $1.03$ seconds. Now, if the normal model is appropriate then, we have,

  $\begin{array}{l}78.21-0.5=77.71\\ 78.21+0.5=78.71\\ z=\frac{77.71-78.21}{1.03}\\ =-0.49\\ z=\frac{78.21-77.71}{1.03}\\ =0.49\end{array}$

And thus the percent of the times should be within $0.5\text{ second of 78}\text{.21}$ is calculated as:

  $\begin{array}{l}=normalcdf(-0.49,0.49)\\ =37.59\%\end{array}$

(b)

To determine

To find out what percent of the times actually falls within this interval.

Answer to Problem 22RE

  $16\%$ .

Explanation of Solution

In the question, it is given the table showing the top women’s skating times. Thus, we know that the mean finishing time was $78.21$ seconds and the standard deviation is $1.03$ seconds. Thus, we have,

  $77.71\le 78.09,78.35,78.59,78.65\le 78.71$

Therefore, the percent of the times actually falls within this interval is calculated as:

  $\begin{array}{l}=\frac{4}{25}\\ =16\%\end{array}$

(c)

To determine

To explain the discrepancy between part (a) and (b).

Explanation of Solution

In the question, it is given the table showing the top women’s skating times. Thus, we know that the mean finishing time was $78.21$ seconds and the standard deviation is $1.03$ seconds. Thus, the times do not actually follow a normal model. The slower time is barely one standard deviation from the mean. Thus, that is why there is discrepancy between part (a) and (b).