Question

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,300 pounds and the standard deviation is 170 pounds. Assume that the population follows the normal distribution. Thirty trucks are randomly selected and weighed.

Within what limits will 99 percent 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 = = 5300

standard deviation = = 170

= 5300

= / n = 170 / 30 = 31.0376

middle 99% of score is

P(-z < Z < z) = 0.99

P(Z < z) - P(Z < -z) = 0.99

2 P(Z < z) - 1 = 0.99

2 P(Z < z) = 1 + 0.99 = 1.99

P(Z < z) = 1.99 / 2 = 0.995

P(Z < 2.58) = 0.995

z  ± 2.58 (see the probability 0.995 in standard normal (Z) table corresponding value is 2.58 )

Using z-score formula  

x= z * +

x= -2.58*31.0376+5300

x= 5219.9

z = 2.58

Using z-score formula  

= z * +

= 2.58*31.0376+5300

= 5380.1