Question

At the U.S. Open Tennis Championship a statistician keeps track of every serve that a player hits during the tournament. The statistician reported that the mean serve speed was 100 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 serve speeds was mound- shaped and symmetric. What percentage of the player's serves were between 115 mph and 145 mph

Answer 1

Answer:

15.74% of the player's serves were between 115 mph and 145 mph

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 100, \sigma = 15$

What percentage of the player's serves were between 115 mph and 145 mph

This is the pvalue of Z when X = 145 subtracted by the pvalue of Z when X = 115.

X = 145

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

$Z = \frac{145 - 100}{15}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

X = 115

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

$Z = \frac{115 - 100}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9987 - 0.8413 = 0.1574

15.74% of the player's serves were between 115 mph and 145 mph