91Ó°ÊÓ

Population Mean $\mathrm{μ}$and Population s.d., $\mathrm{σ}$And the Population variable is normally distributed.

The distribution of $\stackrel{¯}{x}$is also normally

distributed with mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$and S.D ${\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}$, $n=$Sample size.

For any sample size the distribution of $\stackrel{¯}{x}$is always normal but the exact form of the normal distribution i.e.the parameter (S.D$\left.{\mathrm{σ}}_{\stackrel{¯}{B}}\right)$ of the distribution varies for different sample size, since the S.D of $\stackrel{¯}{X},{\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}$. 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 $\stackrel{¯}{x}={\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$

And standard deviation of$\stackrel{¯}{x}={\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}},n=Samplesize.$

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 $\mathrm{μ}$and standard deviation of the sample mean is equal to or approximately equal to $\frac{\mathrm{σ}}{\sqrt{n}}$For $*$SRSWR sampling ${\mathrm{σ}}_{\stackrel{¯}{P}}=\frac{\mathrm{σ}}{\sqrt{n}}$

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

Therefore $\frac{n}{N}â–¡0$since N is large.

$∴$For SRSWOR,${\mathrm{σ}}_{\stackrel{¯}{x}}=\sqrt{\frac{N-n}{N-1}}\frac{\mathrm{σ}}{\sqrt{n}}$

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

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

$∴{\mathrm{σ}}_{\stackrel{¯}{x}}=\sqrt{\frac{1-\frac{n}{N}}{1-\frac{1}{N}}·\frac{\mathrm{σ}}{\sqrt{n}}}$

role="math" localid="1652209872404" $\sqrt{\frac{1-0}{1-0}·\frac{\mathrm{σ}}{\sqrt{n}}}\left[âˆ\mu \frac{n}{N}â–¡\frac{1}{N}â–¡0asislargeN\right]$

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

So, standard deviation of $\stackrel{¯}{X}={\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}.$

Simple random sampling with replacement = SRSWR

** Simple random sampling without replacement =SRSWOR