91Ó°ÊÓ

The question itself says that the distribution is mound-shaped and symmetric, so we will use the empirical rule to find the percentage of serves between 115 mph and 145 mph.

According to the empirical rule, 95% of the observations fall between $\stackrel{¯}{\mathrm{x}}Â\pm 2\mathrm{s}$, and 100% fall between $\stackrel{¯}{\mathrm{x}}Â\pm 3\mathrm{s}$. Because 115 and 145 are greater than 100, both will lie to the right of the mean. Therefore we will only look at the positive side.

If s = 15, $\stackrel{¯}{x}+15=115$ is the first standard deviation from the mean. 2^nd deviation from the mean would be $\stackrel{¯}{x}+2s=100+2\left(15\right)=130$ . And the 3^rd standard deviation from the mean would be $\stackrel{¯}{x}+3s=100+3\left(15\right)=145$.

The graph is given below:

Looking at the graph, we say that 33.32% of players’ serves were between 115 and 145.

$$\begin{gathered} \text{z-score=}\frac{\text{χ}−{\displaystyle \stackrel{-}{\mathrm{χ}}}}{\text{s}} \\ \text{Let'sfindthez-scorefor50mph,} \\ \text{z-score=}\frac{\text{50}−\text{100}}{\text{15}} \\ \text{=}−\text{3.33} \end{gathered}$$

Because the value of the z-score is beyond ±3, 50 mph is an outlier.

z-score of 80 mph,

$\begin{array}{c}\text{z-score=}\frac{\text{80}−\text{100}}{\text{15}}\\ \text{=}−\text{1.33}\end{array}$

80 mph is not an outlierbecause it falls between ±3.

z-score of 105 mph,

$\begin{array}{c}\text{z-score=}\frac{\text{105}−\text{100}}{\text{15}}\\ \text{=0.33}\end{array}$

105 mph is not an outlier.

If the shape of the distribution is not known, we use Chebyshev’s rule. According to which, 75% of observations lie between $\stackrel{¯}{x}Â\pm 2s$ , and 89% lie between $\stackrel{¯}{x}Â\pm 3s$. As 70 < 10, we will calculate the left side of the mean.

$\stackrel{¯}{x}-s=100-15=85$

$\stackrel{¯}{x}-2s=100-2\left(15\right)=70$

Again we will make a distribution graph to make it easier to understand.

The graph is given below:

Therefore, 50 – 37.5 = 12.5%.

12.5% of players’ serves lie below 70 mph.