Question

extra example 2: vicki teaches an aerobics class at a local gym. each student in her class recorded how many hours they spent working out for a month. they also recorded how much weight they lost.
hours spent working out: 4, 5, 8, 9, 9, 13, 16, 21, 26, 28, 31
number of pounds lost: 5, 6, 8, 6, 10, 10, 11, 15, 13, 17, 19
a) find an equation for a line of fit for the data using a scatter plot.
b) use the equation to predict how many hours of working out it would take to lose 25 pounds.

Explanation:

Part (a)

Step 1: Plot the Data Points

First, we identify the independent variable (hours spent working out, \( x \)) and the dependent variable (pounds lost, \( y \)). The data points are: \((4, 5)\), \((5, 6)\), \((8, 8)\), \((9, 6)\), \((9, 10)\), \((13, 10)\), \((16, 11)\), \((21, 15)\), \((26, 13)\), \((28, 17)\), \((31, 19)\). We plot these points on a scatter plot with \( x \)-axis as hours and \( y \)-axis as pounds lost.

Step 2: Choose Two Points for Line of Fit

Looking at the scatter plot, we can choose two points that seem to lie close to the trend. Let's pick \((4, 5)\) and \((31, 19)\).

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

The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Substituting \( (x_1,y_1)=(4,5) \) and \( (x_2,y_2)=(31,19) \), we get:

$$ m=\frac{19 - 5}{31 - 4}=\frac{14}{27}\approx0.5185 $$

Step 4: Find the Equation of the Line

Using the point - slope form \( y - y_1=m(x - x_1) \) with \( (x_1,y_1)=(4,5) \) and \( m=\frac{14}{27} \):

$$ y - 5=\frac{14}{27}(x - 4) $$

Simplify to slope - intercept form (\( y=mx + b \)):

$$ y=\frac{14}{27}x-\frac{56}{27}+5=\frac{14}{27}x-\frac{56}{27}+\frac{135}{27}=\frac{14}{27}x+\frac{79}{27}\approx0.5185x + 2.926 $$

(Note: Another pair of points could be chosen, for example, \((8,8)\) and \((28,17)\). Then \( m=\frac{17 - 8}{28 - 8}=\frac{9}{20} = 0.45 \), and using point - slope with \((8,8)\): \( y - 8=0.45(x - 8)\), \( y=0.45x-3.6 + 8=0.45x + 4.4 \). The line of fit can vary slightly depending on the points chosen, but it should approximate the trend of the data.)

Part (b)

Step 1: Substitute \( y = 25 \) into the Line of Fit Equation

Using the equation we found in part (a), let's use \( y = 0.45x+4.4 \) (using the second line of fit we calculated as an example, the result will be close for different reasonable lines of fit). Substitute \( y = 25 \):

$$ 25=0.45x + 4.4 $$

Step 2: Solve for \( x \)

Subtract 4.4 from both sides:

$$ 25-4.4=0.45x $$

$$ 20.6 = 0.45x $$

Divide both sides by 0.45:

$$ x=\frac{20.6}{0.45}\approx45.78 $$

If we use the first line of fit \( y=\frac{14}{27}x+\frac{79}{27} \), substitute \( y = 25 \):

$$ 25=\frac{14}{27}x+\frac{79}{27} $$

Multiply both sides by 27:

$$ 675 = 14x+79 $$

Subtract 79 from both sides:

$$ 596=14x $$

$$ x=\frac{596}{14}\approx42.57 $$

A more accurate way is to use the two - point formula with a better - fitting line. Let's recalculate the slope using the mean of \( x \) and mean of \( y \).

The mean of \( x \) values: \( \bar{x}=\frac{4 + 5+8 + 9+9+13+16+21+26+28+31}{11}=\frac{170}{11}\approx15.45 \)

The mean of \( y \) values: \( \bar{y}=\frac{5 + 6+8 + 6+10+10+11+15+13+17+19}{11}=\frac{120}{11}\approx10.91 \)

We can also use the formula for the slope of the line of best fit \( m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2} \)

But for a simple line of fit (not the best fit), using the two points \((4,5)\) and \((31,19)\) with \( y=mx + b \), we have:

From \( 5 = 0.5185\times4 + b \), \( 5=2.074 + b \), \( b = 2.926 \), so \( y = 0.5185x+2.926 \)

Substitute \( y = 25 \):

\( 25=0.5185x+2.926 \)

\( 0.5185x=25 - 2.926=22.074 \)

\( x=\frac{22.074}{0.5185}\approx42.57 \)

The approximate number of hours is around 43 - 46 hours (depending on the line of fit chosen).

Part (a) Answer:

An example of a line of fit equation is \( y\approx0.52x + 2.93 \) (or other equations based on the chosen points, e.g., \( y = 0.45x+4.4 \))

Part (b) Answer:

Approximately 43 - 46 hours (for example, if we use \( y = 0.45x + 4.4 \), \( x\approx45.8\approx46 \) hours; if we use \( y=0.52x + 2.93 \), \( x\approx42.6\approx43 \) hours)

Answer:

Step 1: Substitute \( y = 25 \) into the Line of Fit Equation

Using the equation we found in part (a), let's use \( y = 0.45x+4.4 \) (using the second line of fit we calculated as an example, the result will be close for different reasonable lines of fit). Substitute \( y = 25 \):

$$ 25=0.45x + 4.4 $$

Step 2: Solve for \( x \)

Subtract 4.4 from both sides:

$$ 25-4.4=0.45x $$

$$ 20.6 = 0.45x $$

Divide both sides by 0.45:

$$ x=\frac{20.6}{0.45}\approx45.78 $$

If we use the first line of fit \( y=\frac{14}{27}x+\frac{79}{27} \), substitute \( y = 25 \):

$$ 25=\frac{14}{27}x+\frac{79}{27} $$

Multiply both sides by 27:

$$ 675 = 14x+79 $$

Subtract 79 from both sides:

$$ 596=14x $$

$$ x=\frac{596}{14}\approx42.57 $$

A more accurate way is to use the two - point formula with a better - fitting line. Let's recalculate the slope using the mean of \( x \) and mean of \( y \).

The mean of \( x \) values: \( \bar{x}=\frac{4 + 5+8 + 9+9+13+16+21+26+28+31}{11}=\frac{170}{11}\approx15.45 \)

The mean of \( y \) values: \( \bar{y}=\frac{5 + 6+8 + 6+10+10+11+15+13+17+19}{11}=\frac{120}{11}\approx10.91 \)

We can also use the formula for the slope of the line of best fit \( m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2} \)

But for a simple line of fit (not the best fit), using the two points \((4,5)\) and \((31,19)\) with \( y=mx + b \), we have:

From \( 5 = 0.5185\times4 + b \), \( 5=2.074 + b \), \( b = 2.926 \), so \( y = 0.5185x+2.926 \)

Substitute \( y = 25 \):

\( 25=0.5185x+2.926 \)

\( 0.5185x=25 - 2.926=22.074 \)

\( x=\frac{22.074}{0.5185}\approx42.57 \)

The approximate number of hours is around 43 - 46 hours (depending on the line of fit chosen).

Part (a) Answer:

An example of a line of fit equation is \( y\approx0.52x + 2.93 \) (or other equations based on the chosen points, e.g., \( y = 0.45x+4.4 \))

Part (b) Answer:

Approximately 43 - 46 hours (for example, if we use \( y = 0.45x + 4.4 \), \( x\approx45.8\approx46 \) hours; if we use \( y=0.52x + 2.93 \), \( x\approx42.6\approx43 \) hours)