Question

If you know that the average number of hours a sleep a person gets per night is 7 with a standard deviation of 2 in the populating of U.S. adults, what proportion of the population would you expect to sleep between 7 and 9 hours per night? Assume normal distribution. Again, drawing out the distribution may help you

Answer 1

Let X denote the number of hours of sleep a person gets per night. Then $X\sim N(\mu=7,\sigma^{2}=2^{2})$

Required probability = $P(7<X<9)=P(\frac{7-7}{2}<Z<\frac{9-7}{2})=P(0<Z<1)$

$=P(Z<1)-P(Z<0)=0.8413-0.5000=0.3413$