# Using the daily high and low temperature readings at Chicago's O'Hare International Airport for an entire year, a meteorologist made a scatterplot rel

Using the daily high and low temperature readings at Chicago's O'Hare International Airport for an entire year, a meteorologist made a scatterplot relating $y=high$ temperature to $x=low$ temperature, both in degrees Fahrenheit.
After verifying that the conditions for the regression model were met, the meteorologist calculated the equation of the population regression line to be $\left[{\mu }_{y}=16.6+1.02\right]with\left[\sigma =6.6{+}^{\circ }F\right]$.
About what percent of days with a low temperature of ${40}^{\circ }$ F have a high temperature greater than ${70}^{\circ }$ F?

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Viktor Wiley

Step 1
Given:
$\left[{\mu }_{y}=16.6+1.02xri>h\right]$ (Equation population regression line)
$\left[\sigma =6.64\right]$
The average high temperature on days where the low temperature is ${40}^{\circ }$ F according to the population regression line can be found by replacing 2 in the regression line equation by 40 and evaluating.
$\left[{\mu }_{y}=16.6+1.02\left(40\right)=16.6+40.8=57.4\right]$
Thus the mean is 57.4 and the standard deviation is 6.64.
Since the conditions are met, the response y varies according to a Normal distribution.
The z-score is the value decreased by the mean, divided by the standard deviation.
$\left[z=\frac{x-\mu }{\sigma }=\frac{70-57.4}{6.64}\approx 1.90\right]$
Determine the corresponding probability using the normal probability table in the appendix. $\left[P\left(Z<1.90\right)\right]$ is given in the row starting with 1.9 and in the column starting with .00 of the standard normal probability table in the appendix.
$P\left(X>70\right)=P\left(Z>1.90\right)$
$=1-P\left(Z<1.90\right)$
$=1-0.9713$
$=0.0287$
$=2.87\mathrm{%}$
Thus about 2.87% of the days with a low temperature of ${40}^{\circ }$ F are expected to have a high temperature that is greater than ${70}^{\circ }$ F.
Result: 2.87%