The number of hours adults sleep per night is normally distr

The number of hours adults sleep per night is normally distributed with a mean of 7 hours. Assume that the standard deviation is unknown. If 63% of adults sleep more than 6.7 hours per night, what is the variance?

Solution

P ( Z < x ) = 0.63
Value of z to the cumulative probability of 0.63 from normal table is -0.332
P( x-u/s.d < x - 7/s.d) = 0.63
That is, ( 6.7 - 7/s.d ) = -0.332
--> s.d =    ( 6.7 - 7 ) / - 0.332 = 0.9036
Variance is =    0.9036^2 = 0.81649