If the sampled population has a mean 48 and standard deviation
18, then the mean and the standard deviation for the sampling
distribution of (X-bar) for n = 9 are:
A. 48 and 18
B. 48 and 9
C. 16 and 6
D. 48 and 6
E. 48 and 2
Solution :
Given that ,
mean = = 48
standard deviation = = 18
n = 9
sample distribution of sample mean is ,
=
= 48
sampling distribution of standard deviation
= / n = 18 / 9
= 6
D. 48 and 6
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