Question

A trucking company determined that the distance traveled per truck per year is normally​ distributed, with a mean of 50 thousand miles and a standard deviation of 10 thousand miles. Complete parts​ (a) through​ (b) below.

a. What proportion of trucks can be expected to travel between 37 and 50 thousand miles in a​ year?

The proportion of trucks that can be expected to travel between 37 and 50 thousand miles in a year is__.

​(Round to four decimal places as​ needed.)

b. What percentage of trucks can be expected to travel either less than 30 or more than 65 thousand miles in a​year?

The percentage of trucks that can be expected to travel either less than 30 or more than 65 thousand miles in a year is__.

​(Round to two decimal places as​ needed.)

c. How many miles will be traveled by at least 80​% of the​ trucks?

The number of miles that will be traveled by at least 80​% of the trucks is nothing miles.

​(Round to the nearest mile as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 50

standard deviation = = 10

a) P( 37 < x < 50 ) = P[(37 - 50)/ 10) < (x - ) /  < (50 - 50) / 10) ]

= P(-1.30 < z < 0)

= P(z < 0) - P(z < -1.30)

Using z table,

= 0.5 - 0.0968

= 0.4032

b) P(x < 30 or x > 65 )

= P[(x - ) / < (30 - 50) / 10] + 1- p P[(x - ) / < (65 - 50) / 10]

= P(z < -2.00) + 1- P(z < 1.50)

= 0.0228 + ( 1 - 0.9332 )

= 0.0228 + 0.0668

= 0.0896

percentage = 8.96%

c) Using standard normal table,

P(Z z) = 80%

= 1 - P(Z z) = 0.80  

= P(Z z) = 1 - 0.80

= P(Z -0.842 ) = 0.20

z = -0.842

Using z-score formula,

x = z * +

x = -0.842 * 10 + 50

x = 41.58

x = 42 thousand miles