Question

The purpose of the questions is to hammer home that the distribution of sampling means for a large number of samples always m
b) Why is the sampling distribution on the right a narrower distribution?
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Answer #1

a)

The distribution of mean on left figure is calculate for small sample size. And the right figure is calculate for large sample size.

We know,

E(\bar{x}) =E(\frac{1}{n}\sum x_{i})=\frac{1}{n}\sum E(x)= \mu

and

V(\bar{x})=V(\frac{1}{n}\sum x_{i})=\frac{1}{n^{2}}\sum {V(x_{i}})=\frac{\sigma^{2} }{n^{2}}n=\frac{\sigma^{2} }{n}

cov(x_{i},x_{j}) = 0 Since the sample is random.

It is clear from the expression of  V(\bar{x}) that the mean computed by larger sample size will have small variance.

Since the spread of the curve of left figure is higher that the right so the variance of the left figure is higher that the right .

i.e mean of left figure is calculate for small sample size. And the right figure is calculate for large sample size.

b)

Since the sample size is large so the variance of the sampling distribution of right figure is small, so the curve in the right is narrower.

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