**Normal Distribution Exercise: Small Starbucks Coffee** The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 365 grams and a standard deviation of 14 grams. Find the weight that corresponds to each event. Use Excel or Appendix C to calculate the z-value. Round your final answers to 2 decimal places. a. Highest 10 percent: [Input Box] b. Middle 50 percent: [Input Box] to [Input Box] c. Highest 80 percent: [Input Box] d. Lowest 10 percent: [Input Box]

Let X denote the weight of a small coffee. Then X~N(365, 14²)

Part a: The value of X is such that

              $$\begin{gathered} P(Z\ge z)=0.10 \\ \Rightarrow P\left(Z\le z\right)=0.90 \end{gathered}$$

 From the normal tables, we have

                 $P(Z\le 1.28)=0.90$

        $$\begin{gathered} \Rightarrow \frac{X-365}{14}=1.28 \\ \Rightarrow X = 382.92 \end{gathered}$$

Part b: For the middle 50%, X is such that

                    $$\begin{gathered} P(-z\le Z\le z)=0.50 \\ \Rightarrow P\left(\right|Z|\le z)=0.50 \end{gathered}$$

    From the normal tables we have

                        $P\left(\right|Z|\le 0.68)=0.50$

                 $$\begin{gathered} \Rightarrow \frac{X-365}{14}=\pm 0.68 \\ \Rightarrow \frac{X-365}{14}=+0.68 \& \frac{X-365}{14}=-0.68 \\ \Rightarrow X=374.52 \& X=355.48 \end{gathered}$$

Hence the middle 50% lie between the values 355.48 and 374.52