Question

A trucking company determined that the distance traveled per truck per year is normally​ distributed, with a mean of 70 thousand miles and a standard deviation of 12 thousand miles. Complete parts​ (a) through​ (c) below.

a. What proportion of trucks can be expected to travel between 53 and 70 thousand miles in a​ year? (Round to four decimal places as​ needed.)

b. What percentage of trucks can be expected to travel either less than 75 or more than 110 thousand miles in a​ year? (Round to four decimal places as needed.)

c. How many miles will be traveled by at least 85​% of the​ trucks? (Round to two decimal places as needed.)

Answer 1

Solution :

Given that ,

mean = = 70

standard deviation = = 12

P(53 < x < 70) = P((53 - 70 / 12) < (x - ) / < (70 - 70) / 12) )

P(53 < x < 70) = P(-1.4167 < z < 0)

P(53 < x < 70) = P(z < 0) - P(z < -1.4167)

P(53 < x < 70) = 0.5 - 0.0783 = 0.9217

Proportion = 0.9217

(b)

P(x < 75) = P((x - ) / < (75 - 70) / 12) = P(z < 0.4167)

P(x < 75) = 0.6616

Probability = 0.6616

P(x > 110) = 1 - P(x < 110)

= 1 - P((x - ) /  < (110 - 70) / 12)

= 1 - P(z < 3.33)

= 1 - 0.9996

= 0.0004

P(x > 110) = 0.0004

Answer = P(x < 75) +  P(x > 110) = 0.6616 + 0.0004 = 0.662 = 66.2%

(c)

P(Z > z) = 85%

1 - P(Z < z) = 0.85

P(Z < z) = 1 - 0.85 = 0.15

P(Z < -1.036) = 0.15

z = -1.036

Using z-score formula,

x = z * +

x = -1.036 * 12 + 70 = 57.57

Answer = 57.57 miles