Answer:
(a) Probability that the sample mean is Greater than 64 is 0.2033 .
Step-by-step explanation:
We are given that a normal population has a mean of 62 and a standard deviation of 12 and we select a random sample of 25.
Let X bar = sample mean
The z score probability distribution for sample mean is ;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 62
[tex]\sigma[/tex] = population standard deviation = 12
n = random sample = 25
So, probability that the sample mean is Greater than 64 = P(X bar > 64)
P(X bar > 64) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{64-62}{\frac{12}{\sqrt{25} } }[/tex] ) = P(Z > 0.83) = 1 - P(Z <= 0.83)
= 1 - 0.79673 = 0.2033 .
