Answer :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 101, \sigma = 15[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{146 - 101}{15}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
X = 116
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{116 - 101}{15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] 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