Question

At a tennis tournament a statistician keeps track of every serve. The statistician reported that the mean serve speed of a particular player was 101 miles per hour​ (mph) and the standard deviation of the serve speeds was 15 mph. Assume that the statistician also gave us the information that the distribution of the serve speeds was bell shaped. What proportion of the​ player's serves are expected to be between 116 mph and 146 ​mph? Round to four decimal places.

Answer 1

Answer:

0.1574 = 15.74% of the​ player's serves are expected to be between 116 mph and 146 ​mph

Step-by-step explanation:

Problems of normally distributed(bell-shaped) samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 101, \sigma = 15$

What proportion of the​ player's serves are expected to be between 116 mph and 146 ​mph?

This is the pvalue of Z when X = 146 subtracted by the pvalue of Z when X = 116. So

X = 146

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{146 - 101}{15}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

X = 116

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{116 - 101}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9987 - 0.8413 = 0.1574

0.1574 = 15.74% of the​ player's serves are expected to be between 116 mph and 146 ​mph