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Suppose the population of all gasoline mileages is normally distributed with mean= 31.5 mpg and std deviation = .0 mpg. Let y denote mileage randomly

Suppose the population of all gasoline mileages is normally distributed

with mean= 31.5 mpg and std deviation = .0 mpg. Let y denote mileage

randomly selected from this population. Find the probabilities

The answer shows the following :P(y 33.4) = P(y (33.4-31.5)/0.8) = P(Z 2.375) = 0.0088 can you explain further please. How does 2.375 equal .0088

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