## Related questions with answers

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 percent of the sample means occur?

Solution

VerifiedThe mean weight of delivery trucks is $\mu=6000$ pounds and the standard deviation is $\sigma=150$ pounds. Sample size of trucks is $n=40$.

Let $\bar{X}$ be the sample mean.

About $95\%$ of the area under the normal curve is within 1.96 standard deviation of the mean i.e $\mu \pm 1.96\sigma/ \sqrt{n}$

$\begin{align*} \mu+1.96\sigma/ \sqrt{n} =6000+1.96(150/\sqrt{40})=6000+1.96 \times 23.73=6046.51 \end{align*}$

and

$\begin{align*} \mu-1.96\sigma/ \sqrt{n} =6000-1.96(150/\sqrt{40})=6000-1.96 \times 23.73=5953.5 \end{align*}$

About $95\%$ of the sample mean lie between $5953.5$ pounds and $6046.5$ pounds.

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