The shape of the distribution of the time required to get an oil change at a 10-
ID: 3132148 • Letter: T
Question
The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.6 minutes, and the standard deviation for oil-change time is 3.1 minutes. To compute probabilities regarding the sample mean using the Central Limit Theorem, what size sample would be required? What is the probability that a random sample of n = 40 oil changes results in a sample mean time of less than 10 minutes? Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 a.m. and 12 p.m. Treating this as a random sample, what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager.Explanation / Answer
a)
We need a sufficiently large sample size. Usually, it is set at n >= 30. [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 10
u = mean = 11.6
n = sample size = 40
s = standard deviation = 3.1
Thus,
z = (x - u) * sqrt(n) / s = -3.264286617
Thus, using a table/technology, the left tailed area of this is
P(z < -3.264286617 ) = 0.0005487 [ANSWER]
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c)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.1
Then, using table or technology,
z = -1.281551566
As x = u + z * s / sqrt(n)
where
u = mean = 11.6
z = the critical z score = -1.281551566
s = standard deviation = 3.1
n = sample size = 40
Then
x = critical value = 10.97184361 MINUTES [ANSWER]
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