Question

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,750 pounds and the standard deviation is 260 pounds. Assume that the population follows the normal distribution. Forty-five trucks are randomly selected and weighed. Within what limits will 99% of the sample means occur?

Answer 1

99% of the sample means will occur between 5,650.20 pounds and 5,849.80.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.575*(260)/(√(45)) = 99.80`

The lower end of the interval is the mean subtracted by M. So it is 5,750 - 99.80 = 5,650.20

The upper end of the interval is the mean added to M. So it is 99.80 + 5,750 = 5,849.80

99% of the sample means will occur between 5,650.20 pounds and 5,849.80.