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 97 miles per hour (mph) and the standard deviation of the serve speeds was 11 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 119 mph and 130 mph

Answer 1

Answer:

0.0215 = 2.15% of the player's serves are expected to be between 119 mph and 130 mph.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The statistician reported that the mean serve speed of a particular player was 97 miles per hour (mph) and the standard deviation of the serve speeds was 11 mph.

This means that $\mu = 97, \sigma = 11$

What proportion of the player's serves are expected to be between 119 mph and 130 mph?

This is the pvalue of Z when X = 130 subtracted by the ovalue of Z when X = 119.

X = 130

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

$Z = \frac{130 - 97}{11}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

X = 119

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

$Z = \frac{119 - 97}{11}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

0.9987 - 0.9772 = 0.0215

0.0215 = 2.15% of the player's serves are expected to be between 119 mph and 130 mph.