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 12 thousand miles. Complete parts​ (a) through​ (c) below. a. nbsp What proportion of trucks can be expected to travel between 34 and 50 thousand miles in a​ year? The proportion of trucks that can be expected to travel between 34 and 50 thousand miles in a year is . 4082. ​(Round to four decimal places as​ needed.) b. nbsp What percentage of trucks can be expected to travel either less than 40 or more than 65 thousand miles in a​ year?

The percentage of trucks that can be expected to travel either less than 40 or more than 65 thousand miles in a year is ______%. ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 50

standard deviation = = 12

(a)

The proportion of trucks can be expected to travel between 34 and 50 thousand miles in a​ year is,

P(34 < x < 50) = P((34 - 50 / 12) < (x - ) / < (50 - 50) / 12) )

P(34 < x < 50) = P(-1.33 < z < 0)

P(34 < x < 50) = P(z < 0) - P(z < -1.33)

P(34 < x < 50) = 0.5 - 0.0918 = 0.4082

Proportion = 0.4082

(b)  The percentage of trucks can be expected to travel either less than 40 or more than 65 thousand miles in a​ year is,

P(x < 40) = P((x - ) / < (40 - 50) / 12) = P(z < -0.83)

Using standard normal table,

P(x < 40) = 0.2033 = 20.33% and

P(x > 65) = 1 - P(x < 65)

= 1 - P((x - ) / < (65 - 50) / 12)

= 1 - P(z < 1.25)

= 1 - 0.8944

P(x > 65) = 0.1056 = 10.56%

Answer = 20.33% + 10.56% = 30.89%

The percentage of trucks that can be expected to travel either less than 40 or more than 65 thousand miles in a year is 30.89% .