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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?

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

Let Xˉ\bar{X} be the sample mean.

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

μ+1.96σ/n=6000+1.96(150/40)=6000+1.96×23.73=6046.51\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

μ1.96σ/n=60001.96(150/40)=60001.96×23.73=5953.5\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%95\% of the sample mean lie between 5953.55953.5 pounds and 6046.56046.5 pounds.

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