Question

Hit an ace! Professional tennis player Novak Djokovic hits the ball extremely hard. His first-serve speeds can be modeled by a Normal distribution with a mean of 112 miles per hour (mph) and a standard deviation of 5 mph.

a. How often does Djokovic hit his first serve faster than 120 mph?

b. What percent of Djokovic’s first serves are slower than 100 mph?

c. What proportion of Djokovic’s first serves have speeds between 100 and 110 mph?

Short answer

Part (a) Around $5.48\%$of the time Djokovic hits his first serve faster than $120$mph.

Part (b) Djokovic hits around $0.82\%$of the first serves slower than $100$mph.

Part (c) Around $0.3364$ of Djokovic’s first serves have speeds between $100$ and $110$ mph.

Step-by-step solution

Part (a) Step 1: Given information

First serve speed, x = 120 mph

Mean speed, μ = 112 mph

Standard deviation, σ = 5 mph

Part (a) Step 2: Concept

The formula used:$z=\frac{x-\mu }{\sigma }$

Part (a) Step 3: Calculation

Calculate the $z-$score,

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

To find the equivalent probability, use the normal probability table in the appendix.

The usual normal probability table for $P(z<1.60)$has a row that starts with $1.6$and a column that starts with$.00$

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

Therefore,

Around $5.48\%$that time Djokovic hits his first serve faster than $120$mph.

Part (b) Step 1: Calculation

Calculate the z − score,

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

To find the equivalent probability, use the normal probability table in the appendix. See the standard normal probability table for $P(z<-2.40)$and the row that starts with $-2.4$and the column that starts with$.00$

$$\begin{gathered} P(x<100)=P(z<-2.40) \\ =0.0082 \\ =0.82\% \end{gathered}$$

Therefore,

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

Part (c) Step 1: Calculation

Calculate the $z-$score,

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

Or

$z=\frac{x-\mu }{\sigma }=\frac{110-112}{5}=-2.40$

To find the equivalent probability, use the normal probability table in the appendix.

See the standard normal probability table for $P(z<-2.40)$ and look for the row with $-2.4$ and the column with$.00$

Or

The usual normal probability table for $P(z<-0.40)$has a row that starts with$-0.4$and a column that starts with$.00$

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

Therefore,

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