Question

The mileage of used cars in a dealer shop is normally distributed with a mean of 85,000 miles and standard deviation of 25,000 miles

if we choose three cars what is the probability that all of them have a mileage less than 55,000 miles

Answer 1

Solution :

Given that ,

mean = = 85000

standard deviation = = 25000

n = 3

= 85000

=  / n = 25000/ 3=14433.7567

P( <55000 ) = P[( - ) / < (55000 - 85000) /14433.7567 ]

= P(z <-2.08 )

Using z table

= 0.8330