Question

The mpg (miles per gallon for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. what is the probability that the mpg for a randomly selected compact car would be less than 32?

Answer 1

For this Q:

mean=31
st.dev.=0.8
x=32

Calculate z-score:
z=(x-m)/s = (32-31)/0.8=1.25

This z-score value corresponding to the probability of 0.8944

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