Answer :
The probability that a compact car's randomly chosen mpg would be higher than 30 is 0.797.
Given data;
A certain compact car's mpg (miles per gallon) has a mean of 31 and a standard deviation of 1.2 and is regularly distributed.
We need to get the probability that the mpg for a randomly selected compact car would be more than 30.
Therefore;
Mean = μ = 31
Standard deviation = σ = 1.2
Let;
P(X > 30) = P(X - 31/1.2 > 30 - 31/1.2)
= P(Z > -0.833)
= 0.797
Hence, the probability that the mpg for a randomly selected compact car would be more than 30 is 0.797.
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