Answer :
Answer:
0.3204 or 32.04%
Step-by-step explanation:
Population mean weight (μ) = 14.6 pounds
Standard deviation (σ) =3 pounds
Assuming a normal distribution, for any given weight 'X' the correspondent z-score is determined by:
[tex]z=\frac{X-\mu}{\sigma} \\[/tex]
For X= 16 pounds:
[tex]z=\frac{16-14.6}{3}\\z=0.4667[/tex]
A z-score of 0.4667 is equivalent to the 67.96-th percentile of the distribution, the probability of a ball being discarded is:
[tex]P(z>0.4667) = 1- 0.6796 = 0.3204[/tex]
The probability that a randomly selected bowling ball is discarded due to being too heavy to use is 0.3204 or 32.04%