Question 520361
A new fluorescent light to replace incandescent bulbs has come on the market. The box says it has an average life of 8,000 hours with a standard deviation of 750 hours. A random sample of 75 light bulbs is taken. Let X be the mean of the sample. 
a.) What is the probability that X ≥ 7900?
z(7900) = (7900-8000)/750 = -0.1333
P(x >= 7900) = P(z > -0.1333) = normalcdf(-0.1333,100) = 0.5530
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b.) What is the probability that 7800 ≤ X ≤ 8100?
Find the z-values of 7800 and of 8100
Then find the probability that z lies between those 2 z-values.
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Ans: 0.1582
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Cheers,
Stan H.