Question

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. nbsp What proportion of trucks can be expected to travel between 66 and 80 thousand miles in a​ year? The proportion of trucks that can be expected to travel between 66 and 80 thousand miles in a year is . 4192. ​(Round to four decimal places as​ needed.) b. nbsp What percentage of trucks can be expected to travel either less than 55 or more than 95 thousand miles in a​ year? The percentage of trucks that can be expected to travel either less than 55 or more than 95 thousand miles in a year is 7.30​%. ​(Round to two decimal places as​ needed.) c. nbsp How many miles will be traveled by at least 85​% of the​ trucks? The amount of miles that will be traveled by at least 85​% of the trucks is nothing miles. ​(Round to the nearest mile as​ needed.)

Answer 1

X N ( 80 , 10)

What proportion of trucks can be expected to travel between 66 and 80 thousand miles in a​ year?

P ( 66 < X < 80 )

Using Central limit theorem  

P ( (66 - 80) / 10 < < (80 - 80)/10 )

P ( -1.4 < Z < 0)

0.5000 - 0.08076 = 0.4192

P ( 66 < X < 80 ) = 0.4192

Part b) What percentage of trucks can be expected to travel either less than 55 or more than 95 thousand miles in a​ year?

Solution :-  

First find the probability of  less than 55

P ( X < 55 ) = P (   < (55 - 80) / 10 )  

P ( Z < -2.5 ) = 0.00621

Now find the probability of more than 95

P ( X > 95 ) = 1 - P ( X < 95)

P ( X < 95) = P ( < ( 95 - 80 ) /10 )

P ( Z < 1.5 ) = 0.93319

P ( X > 95 ) = 1 - P ( X < 95) = 1 - 0.93319 = 0.06681

adding up this two probabilities = 0.00621 + 0.06681 = 0.07302

To find the percentage, multiply probability by 100 = 0.07302 * 100 = 7.30 %

Part C) How many miles will be traveled by at least 85​% of the​ trucks?

Solution :- Need to find the value of X

P ( X > x ) = 85%

P ( X > x ) = 1 -  P ( X < x ) = 1 - 85% = 15% = 0.15

Find the Z score for 0.1500 in statistical table

Z = - 1.03

So X = - 1.03 Standard Deviation below the mean

X =

X = 80 + ( -1.06) * 10

X = 69.7

Truck will travel 69.7 Miles.

To cross check

P ( X > 69.7 ) = 1 - P ( X < 69.7)

P ( X < 69.7) = P ( < ( 69.7 - 80 ) /10 )\

P ( Z < - 1.03) = 0.15151

P ( X > 69.7 ) = 1 - P ( X < 69.7) = 1 - 0.15151 = 0.84849

If we take the percentage we will get 0.84849 * 100 = 84.85%   85%