Homework Sourseccording to the Sleep Foundation the average nights sleep is
ccording to the Sleep Foundation, the average night\'s sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .4 hours and that the probability distribution is normal.
What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?
What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?
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)?
Solution
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 8
u = mean = 6.8
s = standard deviation = 0.4
Thus,
z = (x - u) / s = 3
Thus, using a table/technology, the right tailed area of this is
P(z > 3 ) = 0.001349898 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 6
u = mean = 6.8
s = standard deviation = 0.4
Thus,
z = (x - u) / s = -2
Thus, using a table/technology, the left tailed area of this is
P(z < -2 ) = 0.022750132 [answer]
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c)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 7
x2 = upper bound = 9
u = mean = 6.8
s = standard deviation = 0.4
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.5
z2 = upper z score = (x2 - u) / s = 5.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.691462461
P(z < z2) = 0.999999981
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.30853752 or 31% [answer]
