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. \(\displaystyle{\left[{P}{\left({Z}{<}{1.90}\right)}\right]}\)</span> 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

=1-0.9713

=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%

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. \(\displaystyle{\left[{P}{\left({Z}{<}{1.90}\right)}\right]}\)</span> 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

=1-0.9713

=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%