Question

A racing car consumes a mean of 104 gallons of gas per race with a variance of 9. if 50 racing cars are randomly selected what is the probability that the sample mean would be greater than 102.9 gallons round your answer to four decimal places

Answer 1

Solution :

Given that,

mean = = 104

variance = 2 = 9

standard deviation = = 3

n=50

= =104

= / n = 3/ 50 = 0.42

P( >102.9 ) = 1 - P( <102.9 )

= 1 - P[( - ) / < (102.9-104) / 0.42]

= 1 - P(z <-2.62 )

Using z table

= 1 - 0.0044

= 0.9956

probability= 0.9956