1. The accumulated miles between repairs for vehicle engines is 25,000 miles wit

Question

1. The accumulated miles between repairs for vehicle engines is 25,000 miles with a standard deviation of 2200 miles. The accumulated miles, which have been recorded over time, follow a normal distribution. Find the probability that an engine you just received will last linger than 27,000 miles. Find the probability that the mean accumulated mileage from a sample of 10 engines a. b. exceeds 27,000 miles. and 3rd quartiles for the accumulated miles between repairs. d. Now, you are looking at vehicle transmissions. The historical data for transmission icates a population mean of 16,000 miles with a standard deviation of 2900 miles. The mileage for transmissions does not follow a normal distribution. Find probability that, in a large train shipment of 40 transmissions, the average mileage for this sample will be less than 15,000 miles. the e. If the average for your transmission sample of are going to declare a stand-down of the workforce to determine what What is the cutoff number of miles for the bottom 10% of your sample average. 40 falls below the bottom 10%, you is going wrong. f. Back to the engines.. . If a single engine is considered a "failure" if it doesn't accumulate at least 22,000 miles between repairs, what is the chance that an engine will fail to meet its anticipated mileage accumulation? g. Given the criteria just stated, what would be the "expected number" of failures in the next 1000 engines that are placed into vehicles?

Explanation / Answer

Sol:

using TI83 cal

first findthe z score for bottom 10%

invnorm(0.10)

=-1.281551567

Z score=-1.281551567

Z=x-mean/sd/sqrt(n)

Need to find X

-1.281551567=x-25000/2200/sqrt(40)

-445.7884=x-25000

x=25000-445.7884

x=24554.21

cutoff number of miles for the bottom 10%=24554.21

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