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 differ from the population mean by more than 1.5 gallons? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 91

standard deviation = = 36

n = 44

= 91

= / n = 36 / 44 = 5.4272

The sample mean would differ from the population mean by more than 1.5 gallons

P( 89.5 < < 92.5 ) = 1 - P((89.5 - 91) /5.4272 <( - ) / < (92.5 - 91) / 5.4272))

= 1 - P(-0.28 < Z < 0.28)

= 1 - ( P(Z < 0.28) - P(Z < -0.28) ) Using standard normal table,  

= 1 - ( 0.6103 - 0.3897 )

= 0.7795

Probability = 0.7795