Question

A racing car consumes a mean of 91 gallons of gas per race with a variance of 36 . If 44 racing cars are randomly selected, what is the probability that the sample mean would be greater than 92.6 gallons? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 91

Variance = ² = 36

Standard deviation = = 6

n = 44

= 91

  =  (/n) = (6 / 44 ) = 0.9045

P ( > 92.6 )

= 1 - P (< 92.6 )

= 1 - P ( -   / ) < ( 92.6 - 91 / 0.9045)

= 1 - P ( z < 1.6 / 0.9045 )

= 1 - P ( z < 1.77)

Using z table

= 1 - 0.9616

= 0.0384

Probability = 0.0384