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 1039872: 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 in less than 5000 hours?
Please help.
thank you
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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 in less than 5000 hours?
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z(5000) = (5000-6500)/750 = -2
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Ans: P(x < 5000) = P(z < -2) = normalcdf(-100,-2) = 0.0228
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Cheers,
Stan H.
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