Trystan Horne

2022-02-12

The manager of the store in the preceding exercise calculated the residual for each point in the scatterplot and made a dotplot of the residuals.

The distribution of residuals is roughly Normal with a mean of $0 and standard deviation of $22.92.

A. What percent of the actual sales amount 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 stores

The distribution of residuals is roughly Normal with a mean of $0 and standard deviation of $22.92.

A. What percent of the actual sales amount 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 stores

golfachukc8

Beginner2022-02-13Added 13 answers

Let x = Sales amount.

Given x ~ Normal ($\mu ,\sigma$ ) where $\mu =\mathrm{\$}0}$ and $\sigma =\mathrm{\$}22.92}$

1)$P(\mu -5<x<\mu +5)$

$=P\left(\frac{\mu -5-\mu}{\sigma}<\frac{x-\mu}{\sigma}<\frac{\mu +5-\mu}{\sigma}\right)$

$=P\left(\frac{-5}{\sigma}<z<\frac{5}{\sigma}\right)$ where $z=\frac{x-\mu}{\sigma}$ ~$N(0,1)$

$=P\left(\frac{-5}{22.92}<z<\frac{5}{22.92}\right)$

$=P(-0.218<z<0.218)$

$=P\left(z<0.218\right)-P(z<-0.218)$

$=P\left(z<0.218\right)-1+P\left(z<0.218\right)$

$=2\times 0.5871-1$ [From Standard Normal Area Table]

$=0.0080$

2)$P(-c<x<c)=0.95$

$=P\left(\frac{-c-\mu}{\sigma}<\frac{x-\mu}{\sigma}<\frac{c-\mu}{\sigma}\right)$

$=P\left(\frac{-c-\mu}{\sigma}<z<\frac{c-\mu}{\sigma}\right)$

On comparing with Normal Distribution,$P(-1.96<z<1.96)=0.95$

$\therefore \frac{-c-\mu}{\sigma}=-1.96\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\frac{c-\mu}{\sigma}=1.96$

$\frac{c+\mu}{\sigma}=1.96\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}c=1.96\times 6)+\mu$

$c=(1.96\times 22.92)-0\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}c=(1.96\times 22.92)+0$

$c=-44.922\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}c=44.922$

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