Question

just the last ones on this

A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 80 thousand miles and a standard deviation of 10 thousand miles. Complete parts (a) through (c) below. a. What proportion of trucks can be expected to travel between 69 and 80 thousand miles in a year? The proportion of trucks that can be expected to travel between 69 and 80 thousand miles in a year is 3643 (Round to four decimal places as needed.) b. What percentage of trucks can be expected to travel either less than 60 or more than 95 thousand miles in a year? The percentage of trucks that can be expected to travel either less than 60 or more than 95 thousand miles in a year is 8.96 % (Round to two decimal places as needed.) c. How many miles will be traveled by at least 75% of the trucks? The amount of miles that will be traveled by at least 75% of the trucks ism iles. (Round to the nearest mile as needed.)

Answer 1

1)

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 80

S.d = 10

C)

From z table, P(z>0.67) = 75%

So, 0.67 = (x - 80)/10 = 86.7 thousands ~ 87 thousand miles

2)

Mean = 74

S.d = 9

For this exam

Z = (92-74)/9 = 2

Different exam

Z = (71-68)/3 = 1

So, a student is better with grade 92 on this exam because the z value for the grade 92 is 2 and z value for 71 is 1