According to Runzheimer International, a typical business traveler spends an average of $281 per day in Chicago. This cost includes hotel, meals, car rental, and incidentals. A survey of 60 randomly selected business travelers who have been to Chicago on business recently is taken. For the population mean of $281 per day, what is the probability of getting a sample average of more than $267 per day if the population standard deviation is $47?
$Given,\phantom{\rule{0ex}{0ex}}samplesize\left(n\right)=60\phantom{\rule{0ex}{0ex}}mean\left(\mu \right)=281\phantom{\rule{0ex}{0ex}}s\mathrm{tan}darddeviation\left(\sigma \right)=47$
$RequiredprobabilityisP(\overline{x}267)\phantom{\rule{0ex}{0ex}}=P\left(\frac{\overline{x}-281}{{\displaystyle \frac{47}{\sqrt{60}}}}\frac{267-281}{{\displaystyle \frac{47}{\sqrt{60}}}}\right)\phantom{\rule{0ex}{0ex}}=P\left(Z-2.31\right)\phantom{\rule{0ex}{0ex}}=0.9896(fromZ-table)\phantom{\rule{0ex}{0ex}}\mathit{T}\mathit{h}\mathit{e}\mathbf{}\mathit{p}\mathit{r}\mathit{o}\mathit{b}\mathit{a}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{g}\mathit{e}\mathit{t}\mathit{t}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{a}\mathbf{}\mathit{s}\mathit{a}\mathit{m}\mathit{p}\mathit{l}\mathit{e}\mathbf{}\mathit{a}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\phantom{\rule{0ex}{0ex}}\mathit{o}\mathit{f}\mathbf{}\mathit{m}\mathit{o}\mathit{r}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathbf{}\mathbf{\$}\mathbf{267}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{d}\mathit{a}\mathit{y}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{9896}$