91Ó°ÊÓ

The population mean tariff rate for all ethanol railroad-car shipments, and $\left(\mathrm{σ}\right)$be the population standard deviation is $\$1,150$In addition, the sample size $n$is $500$, and the sample mean tariff rate $\left(\stackrel{¯}{x}\right)$is regularly distributed.

Let $\left(\mathrm{σ}\right)$be the population standard deviation for all ethanol railroad-car shipments, and $\mathrm{μ}$be the population mean tariff rate. Furthermore, the sample size is $\left(n\right)$and the sample means tariff rate is evenly distributed.

The average tariff rate in the sample ${\mathrm{μ}}_x=\mathrm{μ}$and sample standard deviation as,

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{1,150}{\sqrt{500}} \\ =\frac{1,150}{22.3607} \\ =51.43 \end{gathered}$$

Calculate the likelihood that the mean tariff rate of $500$randomly selected ethanol railroad-car shipments is within $\$100$of the mean tariff rate of all ethanol railroad-car shipments.

The z-score for $\mathrm{μ}-100$is,

$$\begin{gathered} z=\frac{\mathrm{μ}-100-\mathrm{μ}}{51.43} \\ =\frac{-100}{51.43} \\ =-1.94 \end{gathered}$$

The z-score for $\mathrm{μ}+100$is

$$\begin{gathered} z=\frac{{\mathrm{μ}}_{-}100-\mathrm{μ}}{51.43} \\ =\frac{100}{51.43} \\ =1.94 \end{gathered}$$

To find the area between the z-scores, use Table II: Areas under the standard normal curve.

$1.94$is the area to the left of the $z-$score $0.0262$

$1.94$is the area to the left of the $z-$score $0.9738$

$$\begin{gathered} =0.9738-0.0262 \\ =0.9476 \end{gathered}$$

As a result, the probability that the mean tariff rate of randomly selected ethanol railroad-car shipments is within $\$100$the mean tariff rate of all ethanol railroad-car shipments is $0.9476$