Question

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,650 pounds and the standard deviation is 240 pounds. Assume that the population follows the normal distribution. Forty-five trucks are randomly selected and weighed.

Within what limits will 90% of the sample means occur? (Round your z-value to 2 decimal places and final answers to 1 decimal place.)

Sample means _______ to _______

Answer 1

Given that,

mean = = 5650

standard deviation = = 240

n = 45

= 5650

= / n = 240/45=35.78

middle 90% of score is

P(-z < Z < z) = 0.90

P(Z < z) - P(Z < -z) = 0.90

2 P(Z < z) - 1 = 0.90

2 P(Z < z) = 1 + 0.90 = 1.90

P(Z < z) = 1.90/ 2 = 0.95

P(Z < 1.65) = 0.95

z  ±1.65

Using z-score formula  

= ±z * +   

= ±1.65 *35.78+5650

=5591 , 5709.0

Sample means ___5591____ to ___5709.0____