Question

Please show work and answers with the excel formulas

According to the Sleep Foundation, the average night's sleep is 6.5 hours. Assume the standard deviation is 0.5 hours and thatthe probability distribution is normal.

A. What is the probability that a randomly selected person sleeps more than 8 hours?

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

C).  Doctors suggest getting between 7 and 9 hours of sleep each night? What percentage of the population gets this much sleep?

Answer 1

Solution:

   We are given that: According to the Sleep Foundation, the average night's sleep is 6.5 hours. Assume the standard deviation is 0.5 hours and thatthe probability distribution is normal.

That is: Mean = and Standard Deviation =

Part A) What is the probability that a randomly selected person sleeps more than 8 hours?

P( X > 8 ) = .........?

Using Excel command:

=1-NORM.DIST(x , mean , SD , cumulative)

=1 - NORM.DIST( 8 , 6.5 , 0.5 , TRUE)   Press ENTER

=0.0013499

=0.0013

That is: P( X > 8) = 0.0013

Using formula method:

Find z score:

Thus we get:

P( X > 8) = P( Z > 3.00)

P( X > 8) = 1 - P( Z < 3.00)

Look in z table for z = 3.0 and 0.00 and find area.

P( Z < 3.00) = 0.9987

Thus

P( X > 8) = 1 - P( Z < 3.00)

P( X > 8) = 1 - 0.9987

P( X > 8) = 0.0013

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

P( X < 6 ) = .........?

Using Excel command:
=NORM.DIST( x , mean , SD , cumulative)

=NORM.DIST( 6 , 6.5 , 0.5 , TRUE)

=0.1586553

That is: P( X < 6) = 0.1586553

Now using Formula method:

Find z score:

Thus we get:

P( X < 6) = P( Z < -1.00)

Look in z table for z = -1.0 and 0.00 and find area.

P( Z < -1.00) = 0.1587

Thus

P( X < 6) = P( Z < -1.00)

P( X < 6) = 0.1587

Part c) Doctors suggest getting between 7 and 9 hours of sleep each night.

What percentage of the population gets this much sleep?

That is:

P( 7 < X < 9) = .........?

Using Excel command:

=NORM.DIST( Upper x , mean , SD , Cumulative )-NORM.DIST( Lower x , mean , SD , Cumulative )

=NORM.DIST( 9 , 6.5 , 0.5 ,TRUE ) - NORM.DIST(7 , 6.5 , 0.5 ,TRUE)

=0.158655

That is:

P( 7 < X < 9 ) =0.158655

P( 7 < X < 9 ) = 0.1587

By using Formula method:

Find z score for x = 7 and for x = 9

Thus we get:

P( 7 < X < 9) = P ( 1.00 < Z < 5.00)

P( 7 < X < 9) = P ( Z < 5.00) -P( Z < 1.00)

since z = 5.00 is above z = 3.49 and P(Z < 3.49) =0.9998 hence P( Z < 5.00) = 1.0000 .

and for P( Z < 1.00) look in z table for z =1.0 and 0.00 and find area:

P( Z < 1.00) =0.8413

thus

P( 7 < X < 9) = P ( Z < 5.00) -P( Z < 1.00)

P( 7 < X < 9) = 1.0000 - 0.8413

P( 7 < X < 9) = 0.1587