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# Using the health records of ever student at a high school, the school nurse created a scatterplot relating y= text{height (in centimeters) to} x= text{age (in years).} text{After verifying that the conditions for the regression model were met, the nurse calculated the equation of the population regression line to be} mu_{0}=105 + 4.2x text{with} sigma=7 cm. About what percent of 15-year-old students at this school are taller than 180 cm? # Using the health records of ever student at a high school, the school nurse created a scatterplot relating y= text{height (in centimeters) to} x= text{age (in years).} text{After verifying that the conditions for the regression model were met, the nurse calculated the equation of the population regression line to be} mu_{0}=105 + 4.2x text{with} sigma=7 cm. About what percent of 15-year-old students at this school are taller than 180 cm?

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Scatterplots asked 2020-11-10
Using the health records of ever student at a high school, the school nurse created a scatterplot relating $$\displaystyle{y}=\ \text{height (in centimeters) to}\ {x}=\ \text{age (in years).}$$
$$\displaystyle\text{After verifying that the conditions for the regression model were met, the nurse calculated the equation of the population regression line to be}\ \mu_{{{0}}}={105}\ +\ {4.2}{x}\ \text{with}\ \sigma={7}\ {c}{m}.$$ About what percent of 15-year-old students at this school are taller than 180 cm?

## Answers (1) 2020-11-11
Given: (Equation population regression line): $$\displaystyle\mu_{{{y}}}={105}\ +\ {4.2}{x}$$
$$\displaystyle\sigma={7}$$ The average height of 15-year-old students at this high scool according to the population regression line can be found by replacing x in the regression line equation by 15 and evaluating. $$\displaystyle\mu_{{{y}}}={105}\ +\ {4.2}{\left({15}\right)}={105}\ +\ {63}={168}$$ Thus the mean is 168 and the standard deviation is7. 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. $$\displaystyle{z}=\ {\frac{{{x}\ -\ \mu}}{{\sigma}}}=\ {\frac{{{180}\ -\ {168}}}{{{7}}}}\ \approx\ {1.71}$$ Determine the corresponding probability using the normal probability table in the appendix. $$\displaystyle{P}{\left({Z}\ {<}\ {1.71}\right)}$$</span> is given in the row starding with 1.7 and in the column starting with .01 of the standard normal probability table in the appendix. $$\displaystyle{P}{\left({X}\ {>}\ {180}\right)}={P}{\left({Z}\ {>}\ {1.71}\right)}$$
$$\displaystyle={1}\ -\ {P}{\left({Z}\ {<}\ {1.71}\right)}$$</span>
$$\displaystyle={1}\ -\ {0.9564}$$
$$\displaystyle={0.0436}$$
$$\displaystyle={4.36}\%$$ Thus about $$\displaystyle{4.36}\%$$ of the 15-year-old students at this scool are expected to be taller than 180 cm.

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Using the health records of ever student at a high school, the school nurse created a scatterplot relating y = height (in centimeters) to x = age (in years).
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