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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