Question 975407
The production of cars per day at an assembly plant has a mean of 120.5 and a standard deviation of 6.2.
Find the probability that fewer than 100 cars are produced on a random day
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z(100) = (100-120.5)/6.2 = -20.5/6.2 = -3.3
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P(x < 100) = P(z < -3.3) = normalcdf(-100,-3.3) = 0.00048..
Cheers,
Stan H.
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