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
**21.58%**

(Round to two decimal places as needed.)

c. How many miles will be traveled by at least 75% of the trucks?

The number of miles that will be traveled by at least 75% of the trucks is _______miles.

(Round to the nearest mile as needed.)

Answer #1

Solution :

Given that,

mean = = 70

standard deviation = = 11

a ) P (55 < x < 70 )

P ( 55 - 70 / 11) < ( x - / ) < (70 - 70 / 11)

P ( - 15 / 11 < z < 0 / 11 )

P (-1.36 < z < 0 )

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

Using z table

= 0.5000 - 0.0869

= 0.4131

Probability = 0.4131

b ) P( x < 50 )

P ( x - / ) < ( 50 - 70 / 11)

P ( z < -20 / 11 )

P ( z < -1.82 )

= 0.0344

P (x > 80 )

= 1 - P (x < 80 )

= 1 - P ( x - / ) < ( 80 - 70 / 11)

= 1 - P ( z < 10 / 11 )

= 1 - P ( z < 0.91 )

Probability = 0.0344 +0.1814

=02158 = 21.58%

b ) P( Z > z) = 75%

P(Z > z) = 0.75

1 - P( Z < z) = 0.75

P(Z < z) = 1 - 0.75

P(Z < z) = 0.25

z = - 0.67

Using z-score formula,

x = z * +

x = - 0.67 * 11 + 70

x = 63

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