Question
In: Math
The time required for an automotive center to complete an oil change service on an automobile...
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2.5 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 5% of its customers, how long should it make the guaranteed time limit?
Solutions
milcah answered 1 year agoRelated Solutions
The time required for an automotive center to complete an oil change service on an automobile...
The time required for an automotive center to complete an oil
change service on an automobile approximately follows a normal?
distribution, with a mean of 15 minutes and a standard deviation of
4 minutes. ?(a) The automotive center guarantees customers that the
service will take no longer than 20 minutes. If it does take?
longer, the customer will receive the service for? half-price. What
percent of customers receive the service for? half-price? ?(b) If
the automotive center does not want...
The time required for an automotive center to complete an oil change service on an automobile...
The time required for an automotive center to complete an oil
change service on an automobile approximately follows a normal
distribution, with a mean of 19 minutes and a standard deviation of
3
minutes.
(a) The automotive center guarantees customers
that the service will take no longer than 20
minutes. If it does take longer, the customer will receive the
service for half-price. What percent of customers receive the
service for half-price?
(b) If the automotive center does not want...
The time taken by an automobile mechanic to complete an oil change is random with mean...
The time taken by an automobile mechanic to complete an oil
change is random with mean 29.5 minutes and standard deviation 3
minutes.
a. What is the probability that 50 oil changes take more than
1500 minutes?
b. What is the probability that a mechanic can complete 80 or
more oil changes in 40 hours?
c. The mechanic wants to reduce the mean time per oil change so
that the probability is 0.99 that 80 or more oil changes can...
An automotive repair shop has determined that the average service time on an automobile is 2...
An automotive repair shop has determined that the average
service time on an automobile is 2 hours with a standard deviation
of 30 minutes. A random sample of 36 services is selected. (a) What
is the probability that the sample of 36 will have a mean service
time greater than 115 minutes? (b) Assume the population consists
of 400 services. Determine the standard error of the mean.
The shape of the distribution of the time required to get an oil change at a...
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 21.1 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 35 oil changes between 10 A.M. and 12 P.M.
Treating this as a random? sample, there...
the shape of the distribution of the time required to get an oil change in a...
the shape of the distribution of the time required to
get an oil change in a 15 minute oil change facility is on now.
However, records indicate them me Tom is 16.8 minutes, and the
center deviation is 4.4 minutes.
a.) To compute probabilities regarding the sample mean
using the normal model what size would be required?
b.) what is the probability that a random sample of n
equals 40 oil changes results in a sample mean time less than...
The shape of the distribution of the time required to get an oil change at a...
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 is 11.8 minutes, and the
standard deviation is 4.9 minutes. Complete parts (a) through
(c).
(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. Any sample size could be...
The shape of the distribution of the time required to get an oil change at a...
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 16.8 minutes, and the
standard deviation is 4.3 minutes. Complete parts (a) through (c)
below.
To compute probabilities regarding the sample mean using the
normal model, what size sample would be required? Choose the
required sample size below.
A. Any sample size could be used.
B. The normal model cannot be...
The shape of the distribution of the time required to get an oil change at a...
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 21.1 minutes, and the
standard deviation is 4.9 minutes. Complete parts (a) through
(c).
(a) To compute probabilities regarding the sample mean
using the normal model, what size sample would be
required?
A. The normal model cannot be used if the shape of the
distribution is unknown.
B. The sample size...
The shape of the distribution of the time required to get an oil change at a...
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 4.5 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 35 oil changes between 10 A.M. and
12 P.M. Treating this as a random sample,...
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