Question

example 2: harvey’s fitness class at a local gym collected data about the effect of workout time on weight loss. each participant recorded how many hours they spent working out for a month. they also recorded how much weight they lost.
chart with columns: hours spent working out (3, 5, 8, 9, 9, 14, 17, 20, 24, 24, 28) and number of pounds lost (2, 3, 3, 4, 5, 6, 7, 9, 8, 10, 10)
a) find an equation for a line of fit for the data set.
b) use the equation to predict how many pounds someone will lose if they work out for 40 hours

Explanation:

Part (a)

Step 1: Identify Data Points

Let \( x \) be hours spent working out, \( y \) be pounds lost. Data points: \((3,2),(5,3),(8,3),(9,4),(9,5),(14,6),(17,7),(20,9),(24,8),(24,10),(28,10)\).

Step 2: Calculate Mean of \( x \) and \( y \)

\( \bar{x} = \frac{3 + 5 + 8 + 9 + 9 + 14 + 17 + 20 + 24 + 24 + 28}{11} = \frac{161}{11} \approx 14.64 \)
\( \bar{y} = \frac{2 + 3 + 3 + 4 + 5 + 6 + 7 + 9 + 8 + 10 + 10}{11} = \frac{67}{11} \approx 6.09 \)

Step 3: Calculate Slope (\( m \))

Using formula \( m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \). After calculations (omitted for brevity), \( m \approx 0.35 \) (approximate, using two points like \((3,2)\) and \((28,10)\): \( m = \frac{10 - 2}{28 - 3} = \frac{8}{25} = 0.32 \), closer to 0.33).

Step 4: Find Equation of Line

Using point-slope: \( y - \bar{y} = m(x - \bar{x}) \). Using \( m \approx 0.33 \), \( y - 6.09 = 0.33(x - 14.64) \), simplifying to \( y \approx 0.33x + 1.3 \) (or better fit: using two points \((3,2)\) and \((28,10)\), \( y = \frac{8}{25}x + \frac{26}{25} \) or \( y = 0.32x + 1.04 \)).

Part (b)

Step 1: Use the Line of Fit

From part (a), using \( y = 0.32x + 1.04 \) (or similar fit). Substitute \( x = 40 \):
\( y = 0.32(40) + 1.04 = 12.8 + 1.04 = 13.84 \approx 14 \) pounds (or with \( m = 0.33 \), \( y = 0.33(40) + 1.3 = 14.5 \approx 15 \)).

Part (a) Answer: \( y \approx 0.32x + 1.04 \) (or similar linear fit, e.g., \( y = \frac{8}{25}x + \frac{26}{25} \))

Part (b) Answer: Approximately 14 - 15 pounds (e.g., using \( y = 0.32x + 1.04 \), \( y \approx 13.84 \approx 14 \))

Answer:

Step 1: Use the Line of Fit

From part (a), using \( y = 0.32x + 1.04 \) (or similar fit). Substitute \( x = 40 \):
\( y = 0.32(40) + 1.04 = 12.8 + 1.04 = 13.84 \approx 14 \) pounds (or with \( m = 0.33 \), \( y = 0.33(40) + 1.3 = 14.5 \approx 15 \)).

Part (a) Answer: \( y \approx 0.32x + 1.04 \) (or similar linear fit, e.g., \( y = \frac{8}{25}x + \frac{26}{25} \))

Part (b) Answer: Approximately 14 - 15 pounds (e.g., using \( y = 0.32x + 1.04 \), \( y \approx 13.84 \approx 14 \))