Question

The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown. However, records indicate that the mean time is 21.4 minutes, and the standard deviation is 4.2 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 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value. This will be the goal established by the manager.

There is a 10% chance of being at or below a mean oil-change
time of **(answer here)** minutes? (Round to one
decimal place as needed.)

Answer #1

For the sample of size 45, we can apply Central Limit Theorem

So, the sampling distribution of mean will be normally distributed

Mean time = 14x60 /45 = 18.67 minutes

standard deviation of mean time(standard error) =

=

= 0.626

Let the time below which there is 10% chance of oil-change time be A

P(X < A) = P(Z < (A - mean)/standard error)

P(X < A) = 0.10

P(Z < (A - 18.67)/0.626) = 0.10

From standard normal distribution table, take Z score corresponding to 0.1

(A - 18.67)/0.626 = -1.28

A = 17.9 minutes

There is 10% chance of being at or below a mean oil change time of 17.9 minutes

The shape of the distribution of the time required to get an oil
change at a 15
-minute
oil-change facility is unknown. However, records indicate that
the mean time is 16.1 minutes, and the standard deviation is 4.2
minutes.
C) 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...

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change at a
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Suppose the manager agrees to pay each employee a $50 bonus if
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The shape of the distribution of the time required to get an oil
change at a 15-minute oil-change facility is unknown. However,
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The shape of the distribution of the time required to get an oil
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oil-change facility is unknown. However, records indicate that
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and the standard deviation is
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What is the probability that a random sample of
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15
minutes?
Suppose the manager agrees to pay each employee a $50 bonus if
they...

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change at a 10-minute oil-change facility is unknown. However,
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or equal to 30. Please answer C below:
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The shape of the distribution of the time required to get an oil
change at a 20-minute oil-change facility is unknown. However,
records indicate that the mean time is 21.4 minutes, and the
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results in a sample mean time...

The shape of the distribution of the time is required to get an
oil change at a 15-minute oil-change facility is unknown. However,
records indicate that the mean time is 16.2 minutes, and the
standard deviation is 4.2 minutes. Complete parts (a) through
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(a) To compute probabilities regarding the sample mean using the
normal model, what size sample would be required?
A. The sample size needs to be less than or equal to 30.
B. The sample size needs...

The shape of the distribution of the time required to get an oil
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below.
(a) To compute probabilities regarding the sample mean using the
normal model, what size sample would be required?
Choose the required sample size below.
A.
The sample size needs to be greater than 30.
B.
Any...

The shape of the distribution of the time required to get an oil
change facility is unknown. However, records indicate that the mean
time is 21.8 minutes, and the standard deviation is 4.9
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 am and
12pm.Treating this as a random sample there would be a 10% chance...

Suppose the manager agrees to pay each employee a $50 bonus if
they meet a certain goal. On a typical Saturday, the oil-change
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