Question

According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .4 hours and that the probability distribution is normal. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)? What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)? Doctors suggest getting between 7 and 9 hours of sleep each night. What percentage of the population gets this much sleep (to the nearest whole number)?

Answer 1

Let
`X` be the random variable for the time a given person from the population spends sleeping. With
`X\sim\mathcal N(6.8,0.4^2)` we have

`P(X>8)=P\left((X-6.8)/(0.4)>(8-6.8)/(0.4)\right)=P(Z>3)\approx0.0013`

where
`Z\sim\mathcal N(0,1^2)`.

`P(X\le6)=P\left((X-6.8)/(0.4)\le(6-6.8)/(0.4)\right)=P(Z\le-2)\approx0.0228`

`P(7<X<9)=P\left((7-6.8)/(0.4)<(X-6.8)/(0.4)<(9-6.8)/(0.4)\right)=P(0.5<Z<5.5)\approx0.3085`

Rounded to the nearest whole number, that comes out to about 31%.