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 1051187: 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 7500 hours?



.3340 or 33.40%

.1066 or 10.66%

.0742 or 7.42%

.4082 or 40.82%

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
TI syntax is normalcdf(smaller, larger, µ, σ).
p = normalcdf(7225,7500, 6500,750) = .0756