1.In 2009, the American Time Use Survey found that the typical working American
ID: 3153311 • Letter: 1
Question
1.In 2009, the American Time Use Survey found that the typical working American spends an average of 7.5 on the job each weekday. Suppose the standard deviation was found to be 1.2 hours and that working times are normally distributed. If a working American is selected at random on a weekday, what is the probability that he/she worked less than 8 hours that day?
2. In 2009, the American Time Use Survey found that the average 20-24 year-old sleeps an average of 9.4 hours per day. Suppose the standard deviation was reported as 2.2 hours and that sleep times are normally distributed. If a 20-24 year-old was selected at random, what is the probability that they slept between 7 and 9 hours that day?
Explanation / Answer
1.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 8
u = mean = 7.5
s = standard deviation = 1.2
Thus,
z = (x - u) / s = 0.416666667
Thus, using a table/technology, the left tailed area of this is
P(z < 0.416666667 ) = 0.66153888 [ANSWER]
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