Question

A trucking company determined that the distance traveled per truck per year is normally​ distributed, with a mean of 40

thousand miles and a standard deviation of 11 thousand miles

c. How many miles will be traveled by at least 65% of the trucks?

1d) If the standard deviation is 7 thousand miles, the proportion of trucks that can be expected to travel between 27 and 40 thousand miles in a year is?

2d) If the standard deviation is 7 thousasnd miles, the percentage of trucks that can be expected to travel either less than 30 or more than 60 thousand miles in a year is x?

3d) If the standard deviation is 7 thousand miles, the number of miles that will be traveled by at least 65% of the trucks is x miles?

Answer 1

Solution :

Given that,

mean = = 40

c ) standard deviation = = 11

Using standard normal table,

P( Z > z) = 65%

P(Z > z) = 0.65

1 - P( Z < z) = 0.65

P(Z < z) = 1 - 0.65

P(Z < z) = 0.35

z = -0.38

Using z-score formula,

x = z * +

x = - 0.38 * 11+ 40

x = 35.82

1d) standard deviation = = 7

P (27 < x < 40 )

P ( 27 - 40 / 7) < ( x -  / ) < ( 40 - 40 / 7)

P ( -13 / 7 < z < 0 / 7 )

P (-1.86 < z < 0)

P ( z < 0 ) - P ( z < -1.86 )

Using z table

= 0.5000 - 0.0314

= 0.4686

Probability = 0.4686

2d) P( x < 30 )

P ( x - / ) < ( 30 - 40 / 7)

P ( z < -10 / 7 )

P ( z < -1.43 )

= 0.0764

P (x > 60 )

= 1 - P (x < 60 )

= 1 - P ( x -  / ) < ( 60 - 40 / 7)

= 1 - P ( z < 20 / 7 )

= 1 - P ( z < 2.86 )

Using z table

= 1 - 0.9979

= 0.0021

Probability = 0.0764 +0.0021 = 0.0785 = 7.85%

3d) standard deviation = = 7

Using standard normal table,

P( Z > z) = 65%

P(Z > z) = 0.65

1 - P( Z < z) = 0.65

P(Z < z) = 1 - 0.65

P(Z < z) = 0.35

z = -0.38

Using z-score formula,

x = z * +

x = - 0.38 * 7+ 40

x = 37.34