Question

the right. Answer parts (a) and (b) below. Sleeging Timesf Medical Residents Cick to view.page 1o the table. Hours (a) what is he shortest time spent sleeping rat would sal place a resident in the top 5% of sleepog times? Residents who get at lea«Dhours of sleep are, in the top 5% of sleeping times. (Round to two decimal places as needed.) (b) Betwer, what two values does the middle 50% of the sleep times le? The midde 50% of sleep tres lies between ours on the low end and□hours on the high end. (Round to two decimal places as needed.)

Answer 1

$\mathbf{SOLUTION}$

$\mathbf{given:}\;\mu=6.1,\;\sigma=1.0$

$\mathbf{formula:}\;Z=\frac{X-\mu}{\sigma}$

${\color{Blue} \boldsymbol{a)}}\;top\;5\%\Rightarrow P(Z>?)=0.05$

05 1.64

from z-table, Lookup for z-value corresponding to area 0.05 to the right of the normal curve.

$P(Z>1.64)=0.05$

$1.64=\frac{X-6.1}{1.0}$

$X=6.1+(1.64*1.0)$

$X=\mathbf{{\color{Red} 7.74}}$

Residents who get at least 7.74 hours of sleep are in the top 5% of sleeping times.

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${\color{Blue} \boldsymbol{b)}}\;middle\;50\%\Rightarrow P(?<Z<\;?)=0.50$

$\boldsymbol{Assume\;mean\; to\; be\; symmetrically\; distributed}$

$=0.5-\frac{0.50}{2}=\boldsymbol{0.25}$

from z-table, Lookup for z-value corresponding to area 0.25 to the left of the normal curve.

$P(Z<-0.67)=0.25$

$similarly,$

from z-table, Lookup for z-value corresponding to area 0.25 to the right of the normal curve.

$P(Z>0.67)=0.25$

$\therefore P(-0.67<Z<0.67)=0.50$

$\mathbf{Low\; end}$

$-0.67=\frac{X-6.1}{1.0}$

$X=6.1&plus;(-0.67*1.0)$

$X=\mathbf{{\color{Red} 5.43}}$

$\mathbf{high\; end}$

$0.67=\frac{X-6.1}{1.0}$

$X=6.1&plus;(0.67*1.0)$

$X=\mathbf{{\color{Red} 6.77}}$

The middle 50% of sleep times lies between 5.43 hours on the low end and 6.77 hours on the high end.