Chapter 2.2, Problem 62E

(a)

To determine

Percentage of time that Djokovic hits his first serve faster than 120 mph.

Answer to Problem 62E

Around 5.48% of the time Djokovic hits his first serve faster than 120 mph.

Explanation of Solution

Given Information:

First serve speed, $x=120\text{mph}$

Mean speed, $\mu =112\text{mph}$

Standard deviation, $\sigma =5\text{mph}$

Calculations:

Calculate the z − score,

  $z=\frac{120-112}{5}=\frac{3-5.3}{0.9}=1.60$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with 1.6 and the column that starts with .00 of the standard normal probability table for $P\left(z<1.60\right)$ .

  $\begin{array}{l}P\left(x<120\right)=P\left(z>1.60\right)\\ =1-P\left(z<1.60\right)\\ =1-0.9452\\ =0.0548\\ =5.48\%\end{array}$

Therefore,

Around 5.48% of the time Djokovic hits his first serve faster than 120 mph.

(b)

To determine

Percentage ofDjokovic’s first servesslower than 100 mph.

Answer to Problem 62E

Djokovic hits around 0.82% of the first serves slower than 100 mph.

Explanation of Solution

Given Information:

First serve speed, $x=100\text{mph}$

Mean speed, $\mu =112\text{mph}$

Standard deviation, $\sigma =5\text{mph}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{100-112}{5}=-2.40$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -2.4 and the column that starts with .00 of the standard normal probability table for $P\left(z<-2.40\right)$ .

  $\begin{array}{l}P\left(x<100\right)=P\left(z<-2.40\right)\\ =0.0082\\ =0.82\%\end{array}$

Therefore,

Around 0.82% of the first serves are slower than 100 mph.

(c)

To determine

Proportion of Djokovic’s first serves having speeds between 100 and 110 mph.

Answer to Problem 62E

Around 0.3364 of Djokovic’s first serves have speeds between 100 and 110 mph.

Explanation of Solution

Given Information:

First serve speed, $x=100\text{or}110\text{mph}$

Mean speed, $\mu =112\text{mph}$

Standard deviation, $\sigma =5\text{mph}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{100-112}{5}=-2.40$

Or

  $z=\frac{x-\mu }{\sigma }=\frac{110-112}{5}=-0.40$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -2.4 and the column that starts with .00 of the standard normal probability table for $P\left(z<-2.40\right)$ .

Or

See the row that starts with -0.4 and the column that starts with .00 of the standard normal probability table for $P\left(z<-0.40\right)$ .

  $\begin{array}{l}P\left(100<x<110\right)=P\left(-2.40<z<-0.40\right)\\ =P\left(z<-0.40\right)-P\left(z<-2.40\right)\\ =0.3446-0.0082\\ =0.3364\end{array}$

Therefore,

Djokovic’s hits around 0.3364 of the first serves with speeds between 100 and 110 mph.