Question

The manager of the store in the preceding exercise calculated the residual for each point

In the scatterplot and made a dot plot of the residuals. The distribution of residuals is

roughly Normal with a mean of $\backslash (0$and standard deviation of $\backslash )22.92$.

a. What percent of the actual sales amounts do you expect to be within $\$5$of their

expected sales amount?

b. The middle $95\%$of residuals should be between which two values? Use this

information to give an interval of plausible values for the weekly sales revenue if$5$

linear feet are allocated to the store’s brand of men’s grooming products.

Short answer

(a) Actual sales amount is $17.42\%$.

(b) Residual is between $\pm 44.9232$and sales between$1036.4168-1126.2632$

Step-by-step solution

Part (a) Step 1: Given Information

We need to find the percentage of actual sales amount .

Part (a) Step 2: Explanation

We are given that:

$$\begin{gathered} \mathrm{Mean}\left(\mathrm{\mu }\right)=0 \\ \mathrm{Standard}\mathrm{deviation}\left(\mathrm{\sigma }\right)=22.92 \\ \mathrm{x}=-5\mathrm{to}5 \end{gathered}$$

The value of z score is:

$$\begin{gathered} z=\frac{x-\mu }{\sigma }=\frac{-5-0}{22.92}=-0.22 \\ z=\frac{x-\mu }{\sigma }=\frac{5-0}{22.92}=0.22 \end{gathered}$$

Now ,

$$\begin{gathered} P(-5<x<5)=P(-0.22<z<0.22) \\ P(-5<x<5)=-P(z<-0.22)+P(z<0.22) \\ P(-5<x<5)=0.5871-0.4829 \\ P(-5<x<5)=0.1742 \\ P(-5<x<5)=17.42\% \end{gathered}$$

Part (b) Step 1: Given Information

We need to find the residuals and sales.

Part (b) Step 2: Explanation

We are given that;

$$\begin{gathered} \mathrm{\mu }=0 \\ \mathrm{\sigma }=22.92 \\ \mathrm{middle}\mathrm{of}95\%=(100\%-95\%)/2=2.5\% \\ \mathrm{z}=\pm 1.96 \\ \mathrm{and}\mathrm{we}\mathrm{know}\mathrm{that}, \\ \mathrm{z}=\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }} \\ \pm 1.96=\frac{\mathrm{x}-5.3}{22.92} \\ \mathrm{x}=\pm 44.9232 \end{gathered}$$

Now for sales:

$$\begin{gathered} \mathrm{x}=5 \\ \mathrm{y}=317.940+152.680\times 5 \\ \mathrm{y}=1081.340 \\ \mathrm{we}\mathrm{know}\mathrm{that}, \\ \mathrm{y}=\mathrm{residual}+{\mathrm{y}}^{^} \\ \mathrm{y}=\pm 44.9232+1081.340 \\ \mathrm{y}=1036.4168 \\ \mathrm{y}=1126.2632 \end{gathered}$$

Hence residual between $-\$44.9232to\$44.9232$

And sales between $\$1036.4168to1126.2632$