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

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