Question

Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 58,000 miles and the standard deviation was 1,800 miles. What percent of the trucks logged more than 55,060 but less than 59,280 miles?

(MUST SHOW WORK, TYPED ANSWER PLEASE)

Answer 1

Solution :

Given that,

mean = $\mu$ =58,000 miles

standard deviation = $\sigma$ =1,800 miles

A ) P ( x > 55,060 )

= 1 - P (x < 55,060 )

= 1 - P ( x -  $\mu$/ $\sigma$) < ( 55,060 - 58,000 / 1,800 )

= 1 - P ( z < - 2940 / 1,800 )

= 1 - P ( z < - 1.63)

Using z table

= 1 - 0.0516

= 0.9484

Probability = 94.84%

B ) P( x < 59,280)

P ( x -  $\mu$/ $\sigma$) < ( 59,280 - 58,000 / 1,800 )

P ( z < - 5 / 1,800 )

P ( z < 0.71 )

Using z table

0.7611

Probability = 76.11%