SOLUTION: Suppose a simple random sample of size n = 36 is taken from a population with mean = 64 and standard deviation = 18. a) What are the mean and standard deviation of the sampling

Question 1074944: Suppose a simple random sample of size n = 36 is taken from a population with mean = 64 and standard deviation = 18.
a) What are the mean and standard deviation of the sampling distribution?
b) What is P(x < 62.6)?
I generally can figure this out if given the formulas but I don't know the formulas in this instance.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The mean of a sampling distribution is the mean of the population from which it was taken. mean=64
the sd of a sampling distribution is the sd of the population divided by the square root (n)
This would be 18/sqrt(36)=3. The sampling distribution is N~n=36 (64,3)
The probability p<62.6 is a z
z=(x bar-mean)/s/sqrt (n)
z<(62.6-64)/3
z<-1.4/3
z< -0.467.
p=0.3202
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