After studying the process of changing oil, the shop’s manager has found that th
ID: 3386633 • Letter: A
Question
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
Explanation / Answer
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]
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