Answer:
48 and 4
Step-by-step explanation:
If we assume that [tex]\bar X[/tex] to be the random variable that proceed the normal distribution with mean [tex]\mu[/tex] and a standard deviation [tex]\dfrac{\sigma }{\sqrt{n}}[/tex]
Given that:
mean =48
standard deviation = 16
sample size = 16
The population mean is same as the population mean in sampling distribution that is 48.
The standard deviation of the sampling distribution is therefore calculated as:
standard deviation = [tex]\dfrac{\sigma }{\sqrt{n}}[/tex]
standard deviation = [tex]\dfrac{16 }{\sqrt{16}}[/tex]
standard deviation = [tex]\frac{16}{4}[/tex]
standard deviation = 4
Thus; the mean and the standard deviation of the sampling distribution is 48 and 4
