A certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg. A pizza delivery company buys 43 of these cars. What is the probability that the average mileage of the fleet is greater than 30.7 mph
Solution :
Given that,
mean =
= 31
standard deviation =
= 3
n = 43
= 31
=
/
n = 3 /
43 = 0.4575
P(
> 30.7)
= 1 - P(
< 30.7)
= 1 - P[(
-
) /
< (30.7 - 31) / 0.4575]
= 1 - P(z < -0.66)
Using z table,
= 1 - 0.2546
= 0.7454
Probability = 0.7454
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