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Question # 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

Normal distributions
ANSWERED 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 $$\displaystyle{\left[\mu_{{y}}={16.6}+{1.02}\right]}{w}{i}{t}{h}{\left[\sigma={6.6}+^{\circ}{F}\right]}$$.
About what percent of days with a low temperature of $$\displaystyle{40}^{\circ}$$ F have a high temperature greater than $$\displaystyle{70}^{\circ}$$ F? 2020-12-29

Step 1
Given:
$$\displaystyle{\left[\mu_{{y}}={16.6}+{1.02}{x}{r}{i}{>}{h}\right]}$$ (Equation population regression line)
$$\displaystyle{\left[\sigma={6.64}\right]}$$
The average high temperature on days where the low temperature is $$\displaystyle{40}^{\circ}$$ F according to the population regression line can be found by replacing 2 in the regression line equation by 40 and evaluating.
$$\displaystyle{\left[\mu_{{y}}={16.6}+{1.02}{\left({40}\right)}={16.6}+{40.8}={57.4}\right]}$$
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.
$$\displaystyle{\left[{z}={\frac{{{x}-\mu}}{{\sigma}}}={\frac{{{70}-{57.4}}}{{{6.64}}}}\approx{1.90}\right]}$$
Determine the corresponding probability using the normal probability table in the appendix. $$\displaystyle{\left[{P}{\left({Z}{<}{1.90}\right)}\right]}$$ 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.
$$P(X>70)=P(Z>1.90)$$
$$=1-P(Z<1.90)$$
$$=1-0.9713$$
$$=0.0287$$
$$=2.87\%$$
Thus about 2.87% of the days with a low temperature of $$\displaystyle{40}^{\circ}$$ F are expected to have a high temperature that is greater than $$\displaystyle{70}^{\circ}$$ F.
Result: 2.87%