SOLUTION: We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined

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Question 1039389: We are estimating the spares requirement for a radar power supply. The power supply was designed with a mean (μ) life of 6500 hours. The standard deviation (σ) determined from testing is 750 hours. What is the likelihood that a power supply would fail between 7225 and 7670 hours?
.0444 or 4.44%
.1066 or 10.66%
.3340 or 33.40%
.4406 or 44.06%

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
you would calculate the z-scores.
mean is 6500.
standard deviation is 750

the general formula is:
z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.

your low z-score would be:
z = (7225-6500)/750 = .966666666... which can be rounded to .97.

your high z-score would be:
z = (7670-6500)/750 = 1.56

look up the area to the left of the z-score of .97 and the z-score of 1.56 and subtract the smaller area from the larger area to get:

area between z-score of .97 and 1.56 = .9406 minus.8340 = .1066.

your closest solution is .1066 or 10.66% .