Question

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,800 pounds and the standard deviation is 270 pounds. Assume that the population follows the normal distribution. Fifty trucks are randomly selected and weighed.

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

Answer 1

Solution :

Given that,

= / n = 270 / 50 = 38.1838

Using standard normal table,  

P(Z < z) = 90%

P(Z < 1.28) = 0.90

z = 1.28

Using z-score formula,

= z * + = 1.28 * 38.1838 + 5800 = 5848.9