Question 799350: A government agency checked the gasoline mileage of a particular make of automobile and found the mileages to be normally distributed, with a mean of 28.6 mpg and a standard deviation of 2.3 mpg. For one of these automobiles, what is the probability that the mileage will be
a) at least 30 mpg?
b) between 28 and 32 mpg?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A government agency checked the gasoline mileage of a particular make of automobile and found the mileages to be normally distributed, with a mean of 28.6 mpg and a standard deviation of 2.3 mpg. For one of these automobiles, what is the probability that the mileage will be
a) at least 30 mpg?
z(30) = (30-28.6)/2.3 = 0.6087
P(x < 30) = P(z < 0.6087) = normalcdf(-100,0.6087) = 0.7286
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b) between 28 and 32 mpg?
Find the z-value of 28
Find the z-vaue of 32
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Find the probability of z between those two z-values.
Note: I get the following answer:: 0.5332
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Cheers,
Stan H.
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