Question

a magazine reports that the average bowling score for league bowlers in the united states is 157 with a standard deviation of 12, and that the scores are approximately normally distributed

Answer 1

Given that the mean of 15 bowlers that have been selected at random is distributed normally with mean 157 and std dev of 12
The probability that a random sample of 15 bowlers would have an average score greater than 165 will be:
mean=157
std dev,σ =12
std error=σ/√n=12/√15=3.0984
standardizing xbar to z=(xbar-μ)/(σ/√n)
P(xbar>165)=P(165-157)/3.0984
=P(z>2.582)
using normal probability tables we get:
P(z>2.582)=0.0049

Next we calculate the probability that a random sample of 150 bowlers will have an average score greater than 165.
μ=157
σ=12
std error=12/√150=0.9797=0/98
standardizing the xbar we get:
z=(165-157)/0.98
=P(z>8.165)
from normal table this will give us:
P(z>8.165)=0.00