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 11 thousand miles.

a. What proportion of trucks can be expected to travel between 55 and 70 thousand miles in a​ year?

The proportion of trucks that can be expected to travel between 55 and 70 thousand miles in a year is 0.4131

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

b. What percentage of trucks can be expected to travel either less than 50 or more than 80 thousand miles in a​ year?

The percentage of trucks that can be expected to travel either less than 50 or more than 80 thousand miles in a year is _____

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

Answer 1

Solution :

Given that ,

mean = = 70

standard deviation = = 11

P(55 < x <70 ) = P[(55 -70)/ 11) < (x - ) /  < (70 -70) /11 ) ]

= P( -1.36< z <0 )

= P(z < 0) - P(z <-1.36 )

Using standard normal table

= 0.5 - 0.0869 = 0.4131

Probability = 0.4131

P(x < 50 ) = P[(x - ) / < (50 -70) /11 ]

= P(z < -1.82)

= 0.0344

P(x >80 ) = 1 - p( x< 80)

=1- p [(x - ) / < (80 -70) /11 ]

=1- P(z <0.91 )

= 1 - 0.8186= 0.1814

probability = 0.1814 + 0.0344= 0.2158

The percentage of trucks that can be expected to travel either less than 50 or more than 80 thousand miles in a year is 21.58 %.