Answer:
c. 13.6%
Step-by-step explanation:
Mean weight (μ)= 22.0 ounces
Standard deviation (σ) = 1.0 ounce
Assuming a normal distribution, for any given weight X, the corresponding z-score is found by the following equation:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X = 23.1 ounces:
[tex]z=\frac{23.1-22.0}{1.0}\\ z=1.1[/tex]
According to a z table, a score of 1.1 corresponds to the 86.43th percentile of a normal distribution. Therefore, the percentage of regulation basketballs that weight more than 23.1 ounces is:
[tex]P(X>23.1) = 1-0.8643=0.1357=13.57\%[/tex]
The percentage is 13.6%.