Question

According to the Sleep Foundation, the average night’s sleep is 6.8 hours. Assume the standard deviation is .6 hours and that the probability distribution is normal.

What is the probability that a randomly selected person sleeps 6 hours or less?

Doctors suggest getting between 7 and 8 1/2 hours of sleep each night. What percentage of the population gets this much sleep?

Answer 1

Sol)

Given

Mean = 6.8

S.D = 0.6

a)

Randomly selected person sleeps 6 houts or less

P( X <= 6 ) = P( Z < a )

a = ( X - MEAN ) / S.D

= ( 6 - 6.8 ) / 0.6

= -0.8/0.6

= -1.333333

P( X <= 6 ) = P( Z < - 1.33333 )

= 0.0912

b)

Probability of population getting between 7 and 8.5

P( 7 < X < 8.5 ) = P( a < Z < b )

a = ( X - MEAN ) / S.D

a = ( 7 - 6.8 ) / 0.6

= 0.2 / 0.6

= 0.33333

b = ( X - MEAN ) / S.D

= ( 8.5 - 6.8 ) / 0.6

= 1.7 / 0.6

= 2.83333333

P( 7 < X < 8.5 ) = P( 0.33333 < Z < 2.833333)

= P( Z < 2.833333 ) - P( Z < 0.333333 )

= 0.9977 - 0.6306

= 0.3671

= 36.71%

36.71% of population gets this much sleep