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Aloiza [94]
3 years ago
13

Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard devi

ation 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
Mathematics
1 answer:
Alja [10]3 years ago
8 0

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

Answer:

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

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3 years ago
An SRS of 450 450 high school seniors gained an average of ¯ x = 20 x¯=20 points in their second attempt at the SAT Mathematics
maria [59]

Answer: (15.47, 24.53)

Step-by-step explanation:

We know that the confidence interval for population mean is given by :_

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where n= sample size.

\sigma = standard deviation.

\overline{x}= sample mean.

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Given : n= 450

\overline{x}=20

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Critical value for 95% confidence = z*=1.96     [From z-value table]

Then, the 95% confidence interval will be :-

20\pm (1.96)\dfrac{49}{\sqrt{450}}

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Hence, the 95% confidence interval for the mean change in score μ μ in the population of all high school seniors. : (15.47, 24.53)

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