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

A. Sample means _____ to _____

Answer 1

= 5800 pounds

= 270 pounds

n = 50

For sampling distribution of mean, P( < A) = P(Z < (A - )/)

= = 5800 pounds

=

=

= 38.184

Let the 90% of the sample means occur between M and N

P(X < M) = 0.5 - 0.9/2 = 0.05

P(Z < (M - 5800)/38.184) = 0.05

Take the value of Z corresponding to 0.05 from standard normal distribution table.

(M - 5800)/38.184 = -1.65

M = 5737.0 pounds

Similarly,

(N - 5800)/38.184 = 1.645

N = 5863.0 pounds

90% of the sample means occur from 5737.0 pounds to 5863.0 pounds