Homework SourseAfter studying the process of changing oil the shops manager
After studying the process of changing oil, the shop’s manager has found that the distribution of service times, X, is normal with a mean = 28 minutes and a standard deviation = 5 minutes.
Questions: #1 What proportion of cars will be finished in less than half an hour (30 minutes)?
#2 What is the chance that a randomly selected car will take longer than 40 minutes?
#3 What service time corresponds to the 90th percentile?
#4 The manager wants to be able to service 80 percent of the vehicles within 30 minutes. What must be the mean service time be to accomplish this goal
Solution
1.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 30
u = mean = 28
s = standard deviation = 5
Thus,
z = (x - u) / s = 0.4
Thus, using a table/technology, the left tailed area of this is
P(z < 0.4 ) = 0.655421742 [ANSWER]
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2.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 40
u = mean = 28
s = standard deviation = 5
Thus,
z = (x - u) / s = 2.4
Thus, using a table/technology, the right tailed area of this is
P(z > 2.4 ) = 0.008197536 [ANSWER]
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3.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.9
Then, using table or technology,
z = 1.281551566
As x = u + z * s,
where
u = mean = 28
z = the critical z score = 1.281551566
s = standard deviation = 5
Then
x = critical value = 34.40775783 [ANSWER]
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4.
For a 0.80 left tailed area, the corresponding z score is
z = 0.841621234
Thus,
u = x - z*sigma = 30 - 0.841621234*5 = 25.79189383 [ANSWER]
