A variable of a population is normally distribution with mean $\mu$and standard deviation $\sigma .$

a. Identify the distribution of $\overline{x}.$

b. Does your answer to part (a) depend on the sample size? Explain your answer.

c. Identify the mean and the standard deviation of $\overline{x}.$

d. Does your answer to part (c) depend on the assumption that the variable under consideration is normally distributed? Why or why not?

Population Mean $\mu$and Population s.d., $\sigma$And the Population variable is normally distributed.

The distribution of $\overline{x}$is also normally

distributed with mean ${\mu }_{\overline{x}}=\mu$and S.D ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}$, $n=$Sample size.

For any sample size the distribution of $\overline{x}$is always normal but the exact form of the normal distribution i.e.the parameter (S.D$\left.{\sigma }_{\overline{B}}\right)$ of the distribution varies for different sample size, since the S.D of $\overline{X},{\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}$. So, the choice of the sample size does not affect the normality of sampling mean but affects the shape of the normal distribution.

Mean of $\overline{x}={\mu }_{\overline{x}}=\mu$

And standard deviation of$\overline{x}={\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}},n=Samplesize.$

No, it does not depend on the assumption of Normality of the population variable. Since, for any population distribution, mean of the sample mean is equal to Population mean $\mu$and standard deviation of the sample mean is equal to or approximately equal to $\frac{\sigma }{\sqrt{n}}$For $*$SRSWR sampling ${\sigma }_{\overline{P}}=\frac{\sigma }{\sqrt{n}}$

And $*\star$SRSWOR sampling ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}·\frac{\sigma }{\sqrt{n}}$But in general the sample size $n$is much less than population size $N$i.e. $n<N$

Therefore $\frac{n}{N}\square 0$since N is large.

$\therefore$For SRSWOR,${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\frac{\sigma }{\sqrt{n}}$

$=\sqrt{\frac{\frac{N-n}{N}}{\frac{N-1}{N}}}·\frac{\sigma }{\sqrt{n}}\left[\begin{array}{l}\text{dividing the}\\ \text{Numerator \& denominator by}N\end{array}\right]$

$=\sqrt{\frac{\left(1-\frac{n}{N}\right)}{\left(1-\frac{1}{N}\right)}·\frac{\sigma }{\sqrt{n}}}$

$\therefore {\sigma }_{\overline{x}}=\sqrt{\frac{1-\frac{n}{N}}{1-\frac{1}{N}}·\frac{\sigma }{\sqrt{n}}}$

role="math" localid="1652209872404" $\sqrt{\frac{1-0}{1-0}·\frac{\sigma }{\sqrt{n}}}\left[\because \frac{n}{N}\square \frac{1}{N}\square 0asislargeN\right]$

Therefore, we can see that in case of SRSWOR can also be approximated by$\frac{\sigma }{\sqrt{n}}$.

So, standard deviation of $\overline{X}={\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}.$

Simple random sampling with replacement = SRSWR

** Simple random sampling without replacement =SRSWOR