Question

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,950 pounds and the standard deviation is 130 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed.

 

Within what limits will 95 percent of the sample means occur? (Round your z-value to 2 decimal places and final answers to 1 decimal place.)

 

Answer 1

Given,

Mean =5950 Pounds

Standard deviation = 130 pounds

n=40

The value of Z at 95% confidence levele is = 1.96

Now, The 95% confidence limit can be calculated as:

Confidence limit =

t Zx a/Vn

=

59501.96 x 130/V40

=

(5950 1.96 x 130/V40, 5950 1.96 x 130/V40)

=(5909.7, 5990.3)

Hence,The 95% confidence limit is (5909.7 , 5990.3).