37) According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.
a. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?
P( x > 8)
z = ( x - μ ) / σ
z = ( 8 - 6.8 ) / .7
z = 1.714
P(z > 1.714) = 1-0.9564 = 0.0436
b. What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?
P ( x < 6)
z = ( x - μ ) / σ
z = ( 6 - 6.8 ) / .7
z = -1.14
P( z <-1.14) = 0.1271
c. 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)?
P( 7 < x < 9)
z = ( x - μ ) / σ
z = ( 7 - 6.8 ) / .7
z = 0.29
z = ( x - μ ) / σ
z = ( 9 - 6.8 ) / .7
z = 3.14
P ( 0.29 < z < 3.14) = 0.9992 - 0.6141 = 0.3851
38.51%
According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.
a. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?
b. What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?
c. 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)?
3 answers
A-
Z= 8-6.8/6=1.2/6=0.2
B-
Z= 6-6.8/6=0.8/6= 0.133
C-
Z= 7-6.8/6= 0.2/6= 0.333
Z= 9-6.8/6=2.2/6=0.366
Z= 8-6.8/6=1.2/6=0.2
B-
Z= 6-6.8/6=0.8/6= 0.133
C-
Z= 7-6.8/6= 0.2/6= 0.333
Z= 9-6.8/6=2.2/6=0.366
Sorry, but there are a couple of errors in your calculations.
A-
To find the z-score, we need to use the formula:
z = (x - μ) / σ
So,
z = (8 - 6.8) / 0.7 = 1.71
Using the standard normal distribution table or calculator, we find that the probability of z being greater than 1.71 is 0.0438. Therefore, the probability of a randomly selected person sleeping more than 8 hours is 0.0438, or 4.38% (rounded to two decimal places).
B-
To find the probability of a randomly selected person sleeping 6 hours or less, we need to find the area under the normal distribution curve to the left of 6.
z = (6 - 6.8) / 0.7 = -1.14
Using the standard normal distribution table or calculator, we find that the probability of z being less than -1.14 is 0.1271. Therefore, the probability of a randomly selected person sleeping 6 hours or less is 0.1271, or 12.71% (rounded to two decimal places).
C-
To find the probability of a randomly selected person sleeping between 7 and 9 hours, we need to find the area under the normal distribution curve between the z-scores corresponding to 7 and 9.
z1 = (7 - 6.8) / 0.7 = 0.286
z2 = (9 - 6.8) / 0.7 = 3.14
Using the standard normal distribution table or calculator, we find that the probability of z being between 0.286 and 3.14 is 0.4966. Therefore, the probability of a randomly selected person sleeping between 7 and 9 hours is 0.4966, or 49.66% (rounded to two decimal places).
A-
To find the z-score, we need to use the formula:
z = (x - μ) / σ
So,
z = (8 - 6.8) / 0.7 = 1.71
Using the standard normal distribution table or calculator, we find that the probability of z being greater than 1.71 is 0.0438. Therefore, the probability of a randomly selected person sleeping more than 8 hours is 0.0438, or 4.38% (rounded to two decimal places).
B-
To find the probability of a randomly selected person sleeping 6 hours or less, we need to find the area under the normal distribution curve to the left of 6.
z = (6 - 6.8) / 0.7 = -1.14
Using the standard normal distribution table or calculator, we find that the probability of z being less than -1.14 is 0.1271. Therefore, the probability of a randomly selected person sleeping 6 hours or less is 0.1271, or 12.71% (rounded to two decimal places).
C-
To find the probability of a randomly selected person sleeping between 7 and 9 hours, we need to find the area under the normal distribution curve between the z-scores corresponding to 7 and 9.
z1 = (7 - 6.8) / 0.7 = 0.286
z2 = (9 - 6.8) / 0.7 = 3.14
Using the standard normal distribution table or calculator, we find that the probability of z being between 0.286 and 3.14 is 0.4966. Therefore, the probability of a randomly selected person sleeping between 7 and 9 hours is 0.4966, or 49.66% (rounded to two decimal places).
0 answers