91影视

The standard error is 0.20 with a probability equal to 0.90

Here, ${\mathrm{蟽}}^2$is not known, but the observations will range in value between 30 and 34.

a.

The range of the observations will equal $4\mathrm{蟽}$

$\mathrm{Range}鈮�4\mathrm{蟽}$

$\begin{array}{rcl}\mathrm{Range}& =& \mathrm{Maximum}\mathrm{value}-\mathrm{Minimum}\mathrm{value}\\ 4\mathrm{蟽}& =& 34-30\\ \mathrm{蟽}& =& \frac{4}{4}\\ \mathrm{蟽}& =& 1\end{array}$

The sample size is,

$\begin{array}{rcl}{\mathrm{z}}_{\raisebox{1ex}{$\mathrm{伪}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\right)& =& \mathrm{SE}\\ \mathrm{n}& =& {\left[\frac{\left({\mathrm{z}}_{{\displaystyle \raisebox{1ex}{$\mathrm{伪}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\mathrm{蟽}}{\mathrm{SE}}\right]}^2\\ \mathrm{n}& =& {\left[\frac{{\mathrm{z}}_{0.05}脳1}{0.2}\right]}^2\\ \mathrm{n}& =& {\left[\frac{1.645脳1}{0.2}\right]}^2\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ \mathrm{n}& =& \frac{2.706025}{0.04}\\ \mathrm{n}& =& 67.650635\\ n& 鈮�& 68\end{array}$

Hence, the approximate sample size that will produce the desired accuracy of the estimate is 68.

b.

The range of the observations will equal $6\mathrm{蟽}$

$\mathrm{Range}鈮�6\mathrm{蟽}$

$\begin{array}{rcl}\mathrm{Range}& =& \mathrm{Maximum}\mathrm{value}-\mathrm{Minimum}\mathrm{value}\\ 6\mathrm{蟽}& =& 34-30\\ \mathrm{蟽}& =& \frac{4}{6}\\ \mathrm{蟽}& =& 0.6667\end{array}$

The sample size is,

$\begin{array}{rcl}{\mathrm{z}}_{\raisebox{1ex}{$\mathrm{伪}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\right)& =& \mathrm{SE}\\ \mathrm{n}& =& {\left[\frac{\left({\mathrm{z}}_{{\displaystyle \raisebox{1ex}{$\mathrm{伪}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\mathrm{蟽}}{\mathrm{SE}}\right]}^2\\ \mathrm{n}& =& {\left[\frac{{\mathrm{z}}_{0.05}脳0.6667}{0.2}\right]}^2\\ \mathrm{n}& =& {\left[\frac{1.645脳0.6667}{0.2}\right]}^2\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ \mathrm{n}& =& \frac{1.2028}{0.04}\\ \mathrm{n}& =& 30.07\\ \mathrm{n}& 鈮�& 31\end{array}$

Hence, the approximate sample size that will produce the desired accuracy of the estimate is 31.