91Ó°ÊÓ

The mean gross earnings of all Rolling Stones performances have a $99\%$ confidence interval of $\$2.03$ million to $\$2.51$ million.

The formula used:$n={\left(\frac{Z_{\frac{a}{2}}\mathrm{σ}}{E}\right)}^2$and$n={\left\{\frac{Z_{0.025}×\mathrm{σ}}{E}\right\}}^2$

The $99\%$ confidence interval for all Rolling Stones concerts' mean gross earnings, $\mathrm{μ}$ is $\$2.03$ million to $\$2.51$ million.

$∴$Length of the confidence interval

$=\$2.51$million $-\$2.03$million

$=\$0.48$million

$∴$The margin of error $=\frac{1}{2}×$length of the C.I for $\mathrm{μ}$$=\frac{1}{2}×\$0.48$million

$=\$0.24$million

The margin of error $E=\$0.24$ million for a $99\%$confidence interval of the population mean gross earnings $\mathrm{μ}$indicates that we are $99\%$sure that the error in estimating population mean $\mathrm{μ}$by sample mean $\stackrel{¯}{x}$ is at most $\$0.24$ million.

Or, to put it another way, if we take numerous random samples of size $n$ from a population with a mean of $\mathrm{μ}$we may anticipate around $99\%$of the sample means to depart from $\mathrm{μ}$by at most $\$0.24$million.

The sample size $n$required for a $100(1-\mathrm{Î\pm })\%$confidence interval for $\mathrm{μ}$with a certain margin of error $\left(E\right)$is calculated as follows:

$n={\left(\frac{Z_{\frac{a}{2}}\mathrm{σ}}{E}\right)}^2$

Here confidence level $=95\%$

$$\begin{gathered} =100×0.95\% \\ ∴1-\mathrm{Î\pm }=0.95 \\ ⇒\mathrm{Î\pm }=0.05 \\ ⇒\frac{\mathrm{Î\pm }}{2}=0.025 \\ ∴Z_{\frac{a}{2}}=Z_{0.025}=1.96 \end{gathered}$$

The margin of error, $E=\$0.1$ million

Population S.D $\mathrm{σ}=\$0.5$million

$∴n={\left\{\frac{Z_{0.025}×\mathrm{σ}}{E}\right\}}^2$$={\left\{\frac{1.96×0.5}{0.1}\right\}}^2$$=96.04$$≈97$

$∴$The required sample size is $97$

For the population mean gross earnings $\mathrm{μ}$we need a $95$percent confidence interval.

Given that,

Sample mean $\stackrel{¯}{x}=2.35$

Sample size $n=97$

$95\%$confidence interval is given by,

$\left(\stackrel{¯}{x}-Z_{0.025}\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+Z_{0.025}\frac{\mathrm{σ}}{\sqrt{n}}\right)=(\stackrel{¯}{x}-E,\stackrel{¯}{x}+E)$

Where margin of error $E=Z_{0.025}\frac{\mathrm{σ}}{\sqrt{n}}$

$$\begin{gathered} ∴E=Z_{0.025}\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{1.96×0.5}{\sqrt{97}} \\ =0.0995 \end{gathered}$$

$∴95\%$confidence interval of $\mathrm{μ}$

$$\begin{gathered} =(\stackrel{¯}{x}-E,\stackrel{¯}{x}+E) \\ =(2.35-0.0995,2.35+0.0995) \\ =(2.2505,2.4495) \end{gathered}$$