28.11.2020

Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard deviation the same). How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

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24.06.2023, solved by verified expert
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This question is incomplete, the complete question is;

The owners of Spiffy Lube want to offer their customers a 10-minute guarantee on their standard oil change service. If the oil change takes longer than 10 minutes to complete, the customers is given a coupon for a free oil change at the next visit. Based on past history, the owners believe that the timer required to complete an oil change has a normal distribution with a mean of 8.6 minutes and a standard deviation of 1.2 minutes.

Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard deviation the same). How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

Required change in the mean service time is 7.8988

Step-by-step explanation:

Given the data in the question;

How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

let mean = μ

p( x > 10 ) ≤ (1/25)

p( x > 10 ) ≤ 0.4

p( x-μ / 1.2  > 10-μ / 1.2 ) ≤ 0.4

(10-μ / 1.2 ) ≤ 0.4

(10-μ / 1.2 ) ≥ Suppose management could improve the process, №17886841, 28.11.2020 20:23 ( 0.96 )

(10-μ / 1.2 ≥ 1.751

10-μ  = ≥ 1.751 × 1.2

10-μ  ≤ 2.1012

μ ≤ 10 - 2.1012

μ ≤ 7.8988

Therefore, required change in the mean service time is 7.8988

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Mathematics
Step-by-step answer
P Answered by PhD

This question is incomplete, the complete question is;

The owners of Spiffy Lube want to offer their customers a 10-minute guarantee on their standard oil change service. If the oil change takes longer than 10 minutes to complete, the customers is given a coupon for a free oil change at the next visit. Based on past history, the owners believe that the timer required to complete an oil change has a normal distribution with a mean of 8.6 minutes and a standard deviation of 1.2 minutes.

Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard deviation the same). How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

Required change in the mean service time is 7.8988

Step-by-step explanation:

Given the data in the question;

How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

let mean = μ

p( x > 10 ) ≤ (1/25)

p( x > 10 ) ≤ 0.4

p( x-μ / 1.2  > 10-μ / 1.2 ) ≤ 0.4

(10-μ / 1.2 ) ≤ 0.4

(10-μ / 1.2 ) ≥ q_{norm} ( 0.96 )

(10-μ / 1.2 ≥ 1.751

10-μ  = ≥ 1.751 × 1.2

10-μ  ≤ 2.1012

μ ≤ 10 - 2.1012

μ ≤ 7.8988

Therefore, required change in the mean service time is 7.8988

Mathematics
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P Answered by PhD
Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram
Mathematics
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P Answered by PhD

Cost of 7 gallons=$24.50

Cost of 1 gallon=24.50/7=3.5

Cost of 15 gallons=15*3.5=52.5

Cost of 15 gallons will be $52.5

Mathematics
Step-by-step answer
P Answered by PhD

The answer is in the image 

The answer is in the image 
Mathematics
Step-by-step answer
P Answered by PhD

The solution is in the following image

The solution is in the following image
Mathematics
Step-by-step answer
P Answered by PhD

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

Mathematics
Step-by-step answer
P Answered by PhD
The answer is in the image 

The answer is in the image 

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