Question

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Answer 1

An estimate of the standard deviation of the mean is the standard error of the mean,
\(\large SE=\frac{\sigma}{\sqrt n}\)
So
\(\large SE_1=\frac{\sigma}{\sqrt n}=\frac{\sigma}{4}\), and
\(\large SE_2=\frac{\sigma}{\sqrt n}=\frac{\sigma}{5}\), and
So
P1=\(P(\mu-0.2\sigma<X1<\mu+0.2\sigma )=P(\mu-0.8SE_1<X1<\mu+0.8 SE_1)\)
while
P2=\(P(\mu-0.2\sigma<X2<\mu+0.2\sigma )=
From the normal distribribution, it appears quite clearly that P2>P1.

However, X2 and X1 are calculated from different samples, and hence do not have to be identical. Most of the time, X2 will be closer to the true mean, but nothing guarantees that this would happen. Hence in general they will fall on different parts of the normal distribution curve. If there is a wide deviation between X1 and X2, it is still possible that P1>P2.