A variable of a population has mean $\mu$and standard deviation $\sigma$For a large sample size $n$, fill in the blanks, Justify your answers.

a. Approximately _ $\%$of all possible samples have means within $\sigma /\sqrt{n}$of the population mean, $\mu$.

b. Approximately _ $\%$of all possible samples have means within $2\sigma /\sqrt{n}$of the population mean, $\mu$

c. Approximately _ $\%$of all possible samples have means within $3\sigma /\sqrt{n}$of the population mean, $\mu$

d. Approximately __ $\%$of all possible samples have means within $z_{v/2}$of the population mean, $\mu$

Population mean $\mu$& population S.D $\sigma$

Sample size $n$is large

$\therefore$By using CLT, sample mean $\overline{x}$approximates normal distribution with mean $\mu \&$${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

The formula used: Standard deviation=${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Around $68.26\%$of all feasible samples have means that are within $\frac{\sigma }{\sqrt{n}}$of the population mean $\mu$

Justification: let $z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}$

Then $z~N(0,1)$

Now for standard normal variable $z$

$$\begin{gathered} P(-1\le z\le 1)=.6826 \\ \Rightarrow P\left(-1\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le 1\right)=.6826 \\ \Rightarrow P\left(-{\sigma }_{\overline{x}}\le \overline{x}-\mu \le {\sigma }_{\overline{x}}\right)=.6826 \\ \Rightarrow P\left(\mu -{\sigma }_{\overline{x}}\le \overline{x}\le \mu +{\sigma }_{\overline{x}}\right)=.6826 \\ \text{i.e}P\left(\mu -\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +\frac{\sigma }{\sqrt{n}}\right)=.6826 \end{gathered}$$

Hence the proof of the statement in $\left(a\right)$

The population mean $\mu$is within $\frac{2\sigma }{\sqrt{n}}$of the means of approximately $95.44\%$percent of all possible samples.

Justification:

$$\begin{gathered} P(-2\le z\le 2)=.9544\text{Where}z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}} \\ \Rightarrow P\left(-2\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}\le 2\right)=.9544z~N(0,1) \\ \Rightarrow P\left(-2{\sigma }_{\overline{X}}\le \overline{x}-\mu \le 2{\sigma }_{\overline{X}}\right)=.9544 \\ \Rightarrow P\left(\mu -2{\sigma }_{\overline{x}}\le \overline{x}\le \mu +2{\sigma }_{\overline{X}}\right)=.9544 \\ \Rightarrow P\left(\mu -2\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +2\frac{\sigma }{\sqrt{n}}\right)=.9544 \end{gathered}$$

$99.74\%$of all feasible samples have means that are within $\frac{3\sigma }{\sqrt{n}}$of the population mean $\mu$

Justification: $P(-3\le z\le 3)=.9974z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}$

$$\begin{gathered} z~N(0,1) \\ \Rightarrow P\left(-3\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}\le 3\right)=.9974 \\ \Rightarrow P\left(-3{\sigma }_{\overline{x}}\le \overline{x}-\mu \le 3{\sigma }_{\overline{x}}\right)=.9974 \\ \Rightarrow P\left(\mu -3{\sigma }_{\overline{x}}\le \overline{x}\le \mu +3{\sigma }_{\overline{x}}\right)=.9974 \\ \Rightarrow P\left(\mu -3\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +3\frac{\sigma }{\sqrt{n}}\right)=.9974 \end{gathered}$$

The means of approximately $100(1-\alpha )\%$of all feasible samples are within $z_{\frac{\alpha }{2}}·\frac{\sigma }{\sqrt{n}}$mean $\mu$of the population

Justification:$$\begin{gathered} P\left(-z_{\frac{\alpha }{2}}\le z\le z_{\frac{\alpha }{2}}\right)=1-\alpha \\ \Rightarrow P\left(-z_{\frac{\alpha }{2}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{X}}}\le z_{\frac{\alpha }{2}}\right)=1-\alpha \end{gathered}$$

$$\begin{gathered} \Rightarrow P\left(\mu -z_{\frac{a}{2}}{\sigma }_{\overline{x}}\le \overline{x}\le \mu +z_{\frac{\alpha }{2}}·{\sigma }_{\overline{x}}\right)=1-\alpha \\ \Rightarrow P\left(\mu -\frac{z\alpha }{2}·\frac{\sigma }{\sqrt{n}}\le \overline{x}\le \mu +z_{\frac{\alpha }{2}}·\frac{\sigma }{\sqrt{n}}\right)=1-\alpha \end{gathered}$$