91Ó°ÊÓ

Population mean $\mathrm{μ}$& population S.D $\mathrm{σ}$

Sample size $n$is large

$∴$By using CLT, sample mean $\stackrel{¯}{x}$approximates normal distribution with mean $\mathrm{μ}\&$${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

The formula used: Standard deviation=${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

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

Justification: let $z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}$

Then $z~N(0,1)$

Now for standard normal variable $z$

$$\begin{gathered} P(-1≤z≤1)=.6826 \\ ⇒P\left(-1≤\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{x}}}≤1\right)=.6826 \\ ⇒P\left(-{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}-\mathrm{μ}≤{\mathrm{σ}}_{\stackrel{¯}{x}}\right)=.6826 \\ ⇒P\left(\mathrm{μ}-{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+{\mathrm{σ}}_{\stackrel{¯}{x}}\right)=.6826 \\ \text{i.e}P\left(\mathrm{μ}-\frac{\mathrm{σ}}{\sqrt{n}}≤\stackrel{¯}{x}≤\mathrm{μ}+\frac{\mathrm{σ}}{\sqrt{n}}\right)=.6826 \end{gathered}$$

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

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

Justification:

$$\begin{gathered} P(-2≤z≤2)=.9544\text{Where}z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}} \\ ⇒P\left(-2≤\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤2\right)=.9544z~N(0,1) \\ ⇒P\left(-2{\mathrm{σ}}_{\stackrel{¯}{X}}≤\stackrel{¯}{x}-\mathrm{μ}≤2{\mathrm{σ}}_{\stackrel{¯}{X}}\right)=.9544 \\ ⇒P\left(\mathrm{μ}-2{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+2{\mathrm{σ}}_{\stackrel{¯}{X}}\right)=.9544 \\ ⇒P\left(\mathrm{μ}-2\frac{\mathrm{σ}}{\sqrt{n}}≤\stackrel{¯}{x}≤\mathrm{μ}+2\frac{\mathrm{σ}}{\sqrt{n}}\right)=.9544 \end{gathered}$$

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

Justification: $P(-3≤z≤3)=.9974z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}$

$$\begin{gathered} z~N(0,1) \\ ⇒P\left(-3≤\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤3\right)=.9974 \\ ⇒P\left(-3{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}-\mathrm{μ}≤3{\mathrm{σ}}_{\stackrel{¯}{x}}\right)=.9974 \\ ⇒P\left(\mathrm{μ}-3{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+3{\mathrm{σ}}_{\stackrel{¯}{x}}\right)=.9974 \\ ⇒P\left(\mathrm{μ}-3\frac{\mathrm{σ}}{\sqrt{n}}≤\stackrel{¯}{x}≤\mathrm{μ}+3\frac{\mathrm{σ}}{\sqrt{n}}\right)=.9974 \end{gathered}$$

The means of approximately $100(1-\mathrm{Î\pm })\%$of all feasible samples are within $z_{\frac{\mathrm{Î\pm }}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}$mean $\mathrm{μ}$of the population

Justification:$$\begin{gathered} P\left(-z_{\frac{\mathrm{Î\pm }}{2}}≤z≤z_{\frac{\mathrm{Î\pm }}{2}}\right)=1-\mathrm{Î\pm } \\ ⇒P\left(-z_{\frac{\mathrm{Î\pm }}{2}}≤\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤z_{\frac{\mathrm{Î\pm }}{2}}\right)=1-\mathrm{Î\pm } \end{gathered}$$

$$\begin{gathered} ⇒P\left(\mathrm{μ}-z_{\frac{a}{2}}{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+z_{\frac{\mathrm{Î\pm }}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}\right)=1-\mathrm{Î\pm } \\ ⇒P\left(\mathrm{μ}-\frac{z\mathrm{Î\pm }}{2}·\frac{\mathrm{σ}}{\sqrt{n}}≤\stackrel{¯}{x}≤\mathrm{μ}+z_{\frac{\mathrm{Î\pm }}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}\right)=1-\mathrm{Î\pm } \end{gathered}$$