Question 1026798: A muffler company advertises that you will receive a rebate if it takes longer than 30 minutes to replace a muffler. Experience has shown that the time taken to replace a muffler is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2.5 minutes.
a) What proportion of mufflers take between 22 and 26 minutes to replace?
b) What should be the rebate time of 30 minutes be changed to if the company wishes to provide only 1% of customers with a rebate?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
A muffler company advertises that you will receive a rebate if it takes longer than 30 minutes to replace a muffler. Experience has shown that the time taken to replace a muffler is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2.5 minutes.
a) What proportion of mufflers take between 22 and 26 minutes to replace?
b) What should be the rebate time of 30 minutes be changed to if the company wishes to provide only 1% of customers with a rebate?
Solution:
You will need a normal distribution table, or a calculator or software that provides the same information. Also learn how to use the table, most of which gives the left-tail. (check: P(Z=1)=0.8413447)
Assuming (as given) normal distribution and no distinction between mechanics, then we have
μ=30 minutes, and
σ=2.5 minutes.
recall that
Z=(X-μ)/σ
(a) proportion of mufflers that take 22 and 26 minutes to replace:
P(22≤x≤26)
=P((22-25)/2.5≤Z≤(26-25)/2.5)
=P(-1.2≤Z≤0.4)
=P(0.4)-P(-1.2)
=0.6554-0.1151
=0.5404
(b) New rebate time:
Here we need to find Z such that
P(z≤Z)=0.99
From tables, P(z<2.326348)=0.99, therefore
Z=2.325348=(x-25)/2.5
x=30.81587, or 30 minutes and 49 seconds.
Answer, the rebate time limit should be changed to 30 minutes and 49 seconds.
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