Answer:
13.567%
Step-by-step explanation:
We solve the above question, using z score formula
z = (x-μ)/σ, where
x is the raw score = 23.1 ounces
μ is the population mean = 22.0 ounces
σ is the population standard deviation = 1.0 ounce
More than = Greater than with the sign = >
Hence, for x > 23.1 ounces
z = 23.1 - 22.0/1.0
= 1.1
Probability value from Z-Table:
P(x<23.1) = 0.86433
P(x>23.1) = 1 - P(x<23.1)
P(x>23.1) = 1 - 0.86433
P(x>23.1) = 0.13567
Converting to percentage
= 0.13567 × 100
= 13.567%
Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%
Answer:
458
Step-by-step explanation:
Answer: sometimes
Step-by-step explanation:
