# 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

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.
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
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Step 1
Given:
$\mu =Mean=0$

Middle 95% of residuals
The lower cutoff value of the middle 95% has $\frac{100\mathrm{%}-95\mathrm{%}}{2}=\frac{5\mathrm{%}}{2}=2.5\mathrm{%}$ of the data values below it.
Similarly, the upper cutoff value of the middle 95% has 2.5% of the data values above it and thus $100\mathrm{%}-2.5\mathrm{%}=97.5\mathrm{%}$ of the data values below it.
Let us determine the z-score that corresponds with a probability of 2.5% or 0.025 in the normal probability table of the appendix.
We note that the closest probability is 0.0250 which lies in the row -1.9 and in the column .06 of the normal probability table and thus the corresponding z-score is then $1.9+06=-1.96$.
Similarly, we obtain that the z-score 1.96 corresponds with a probability of 0.9750.
$z=±1.96$
The z-score is the value decreased by the mean, divided by the standard deviation.
$z=\frac{x-\mu }{\sigma }=\frac{x-0}{22.92}$
The two found expressions of the z-score then have to be equal:
$\frac{x-5.3}{22.92}=±1.96$
Multiply each side by 22.92:
$x-0=±1.96\left(22.92\right)$
Add 5.3 to each side:
$x=0±1.96\left(22.92\right)$
Evaluate:
$x=±44.9232$
Thus the middle 95% of the residuals should be between -$44.9232 and$44.9232.
Step 2
Interval sales
$x=5$
$\stackrel{^}{y}=317.940+152.680x$ (result previous exercise)
Let us first determine the predicted value, by evaluating the equation of the
least-squares regression line at $x=5$.
$\stackrel{^}{y}=317.940+152.680\left(5\right)=317.940+763.40=1081.340$
The residual is the difference between the actual y-value (sales) and the predicted y-value:
Residual $=y-\stackrel{^}{y}$
We are interested in the actual sales, thus let us add hat{y} to each side of the previous equation:
y=Residual $+\stackrel{^}{y}$
We know that the residuals are between -$44.9232 and$44.9232.
$y=$Residual $+\stackrel{^}{y}=-\mathrm{}44.9232+\mathrm{}1081.340=\mathrm{}1036.4168$
$y=$Residual $+\stackrel{^}{y}=-\mathrm{}44.9232+\mathrm{}1081.340=\mathrm{}1126.2632$
Thus the plausible values for the weekly sales revenue are then between
$1036.4168 and between$1126.2632.
Result:
Residuals: Between -$44.9232 and$44.9232.
Sales:Between $1036.4168 and between$1126.2632.