QUESTION IMAGE
Question
- using the linear best - fit model for the data, what is the predicted running heart rate of a girl whose walking heart rate is 100 bpm? (round your answer to the hundredths place.) type a response
walking heart rate: 66, 72, 74, 76, 80, 84, 86, 88
running heart rate: 128, 136, 134, 138, 142, 146, 148, 152
Step1: Identify Variables
Let \( x \) be Walking Heart Rate (independent variable) and \( y \) be Running Heart Rate (dependent variable). The data points are:
\((66, 128)\), \((72, 136)\), \((74, 134)\), \((76, 138)\), \((80, 142)\), \((84, 146)\), \((86, 148)\), \((88, 152)\)
Step2: Calculate Mean of \( x \) and \( y \)
Mean of \( x \) (\(\bar{x}\)):
Mean of \( y \) (\(\bar{y}\)):
Step3: Calculate Slope (\( m \))
Slope formula: \( m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)
Calculate \( (x_i - \bar{x})(y_i - \bar{y}) \) and \( (x_i - \bar{x})^2 \) for each point:
- For \( (66, 128) \): \( (66 - 78.25)(128 - 140.5) = (-12.25)(-12.5) = 153.125 \); \( (-12.25)^2 = 150.0625 \)
- For \( (72, 136) \): \( (72 - 78.25)(136 - 140.5) = (-6.25)(-4.5) = 28.125 \); \( (-6.25)^2 = 39.0625 \)
- For \( (74, 134) \): \( (74 - 78.25)(134 - 140.5) = (-4.25)(-6.5) = 27.625 \); \( (-4.25)^2 = 18.0625 \)
- For \( (76, 138) \): \( (76 - 78.25)(138 - 140.5) = (-2.25)(-2.5) = 5.625 \); \( (-2.25)^2 = 5.0625 \)
- For \( (80, 142) \): \( (80 - 78.25)(142 - 140.5) = (1.75)(1.5) = 2.625 \); \( (1.75)^2 = 3.0625 \)
- For \( (84, 146) \): \( (84 - 78.25)(146 - 140.5) = (5.75)(5.5) = 31.625 \); \( (5.75)^2 = 33.0625 \)
- For \( (86, 148) \): \( (86 - 78.25)(148 - 140.5) = (7.75)(7.5) = 58.125 \); \( (7.75)^2 = 60.0625 \)
- For \( (88, 152) \): \( (88 - 78.25)(152 - 140.5) = (9.75)(11.5) = 112.125 \); \( (9.75)^2 = 95.0625 \)
Sum of \( (x_i - \bar{x})(y_i - \bar{y}) \):
Sum of \( (x_i - \bar{x})^2 \):
Slope \( m \):
Step4: Calculate Intercept (\( b \))
Using \( \bar{y} = m\bar{x} + b \):
Step5: Linear Equation
The linear best - fit model is \( y = 0.8107x + 77.06 \)
Step6: Predict \( y \) when \( x = 100 \)
Substitute \( x = 100 \) into the equation:
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