Question

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed.Within what limits will 95% of the sample means occur?

Answer 1

Answer:

(5953.52,6046.49)

Step-by-step explanation:

We are given the following in the question:

Mean, $\mu$ = 6,000 pounds

Sample size, n = 40

Alpha, α = 0.05

Standard deviation, σ = 150 pounds

95% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$6000 \pm 1.96(\dfrac{150}{\sqrt{40}} )\\\\ = 6000 \pm 46.4854=\\(5953.5146,6046.4854)\approx (5953.52,6046.49)$

are the limits within which 95% of the sample means occur.