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6. the data in the table to the left gives the weight, in pounds, and b…

Question

  1. the data in the table to the left gives the weight, in pounds, and bmi (body mass index) for six people of the same height.

a) find the line of best fit:
b) sam is the same height as the people in this group. if he has a bmi of 25, estimate his weight.
weight | 150 | 155 | 160 | 165 | 170 | 175
bmi | 20.3 | 21.0 | 21.7 | 22.4 | 23.1 | 23.7

Explanation:

Part a) Line of Best Fit

To find the line of best fit \( y = mx + b \) where \( y \) is BMI and \( x \) is Weight, we follow these steps:

Step 1: Calculate \( \bar{x} \) (mean of Weight) and \( \bar{y} \) (mean of BMI)
  • Weight values (\( x \)): 150, 155, 160, 165, 170, 175

\( \bar{x} = \frac{150 + 155 + 160 + 165 + 170 + 175}{6} = \frac{975}{6} = 162.5 \)

  • BMI values (\( y \)): 20.3, 21.0, 21.7, 22.4, 23.1, 23.7

\( \bar{y} = \frac{20.3 + 21.0 + 21.7 + 22.4 + 23.1 + 23.7}{6} = \frac{132.2}{6} \approx 22.0333 \)

Step 2: Calculate the slope \( m \)

The formula for \( m \) is:

$$ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} $$

First, compute \( (x_i - \bar{x})(y_i - \bar{y}) \) and \( (x_i - \bar{x})^2 \) for each data point:

\( x_i \)\( y_i \)\( x_i - \bar{x} \)\( y_i - \bar{y} \)\( (x_i - \bar{x})(y_i - \bar{y}) \)\( (x_i - \bar{x})^2 \)
15521.0-7.5-1.0333\( (-7.5)(-1.0333) \approx 7.75 \)\( (-7.5)^2 = 56.25 \)
16021.7-2.5-0.3333\( (-2.5)(-0.3333) \approx 0.833 \)\( (-2.5)^2 = 6.25 \)
16522.42.50.3667\( (2.5)(0.3667) \approx 0.917 \)\( (2.5)^2 = 6.25 \)
17023.17.51.0667\( (7.5)(1.0667) \approx 8.000 \)\( (7.5)^2 = 56.25 \)
17523.712.51.6667\( (12.5)(1.6667) \approx 20.833 \)\( (12.5)^2 = 156.25 \)
Step 3: Sum the products and squared differences
  • \( \sum (x_i - \bar{x})(y_i - \bar{y}) \approx 21.666 + 7.75 + 0.833 + 0.917 + 8.000 + 20.833 = 60 \)
  • \( \sum (x_i - \bar{x})^2 = 156.25 + 56.25 + 6.25 + 6.25 + 56.25 + 156.25 = 437.5 \)
Step 4: Calculate slope \( m \)
$$ m = \frac{60}{437.5} \approx 0.1371 $$
Step 5: Calculate intercept \( b \)

Using \( \bar{y} = m\bar{x} + b \):

$$ 22.0333 = 0.1371(162.5) + b $$
$$ 22.0333 \approx 22.28 + b $$
$$ b \approx 22.0333 - 22.28 \approx -0.2467 $$

Thus, the line of best fit is approximately \( \boldsymbol{y = 0.137x - 0.247} \) (or rounded for simplicity, \( y \approx 0.137x - 0.25 \)).

Part b) Estimate Weight for BMI = 25

Using the line of best fit \( y = 0.137x - 0.247 \) (where \( y = 25 \)):

Step 1: Solve for \( x \)
$$ 25 = 0.137x - 0.247 $$
$$ 25 + 0.247 = 0.137x $$
$$ 25.247 = 0.137x $$
$$ x = \frac{25.247}{0.137} \approx 184.3 $$

Thus, Sam’s estimated weight is approximately 184 pounds (rounded to a reasonable whole number).

Final Answers

a) Line of best fit: \( \boldsymbol{y \approx 0.137x - 0.25} \) (or more precisely \( y \approx 0.137x - 0.247 \))
b) Estimated weight: \( \boldsymbol{\approx 184} \) pounds

Answer:

Part a) Line of Best Fit

To find the line of best fit \( y = mx + b \) where \( y \) is BMI and \( x \) is Weight, we follow these steps:

Step 1: Calculate \( \bar{x} \) (mean of Weight) and \( \bar{y} \) (mean of BMI)
  • Weight values (\( x \)): 150, 155, 160, 165, 170, 175

\( \bar{x} = \frac{150 + 155 + 160 + 165 + 170 + 175}{6} = \frac{975}{6} = 162.5 \)

  • BMI values (\( y \)): 20.3, 21.0, 21.7, 22.4, 23.1, 23.7

\( \bar{y} = \frac{20.3 + 21.0 + 21.7 + 22.4 + 23.1 + 23.7}{6} = \frac{132.2}{6} \approx 22.0333 \)

Step 2: Calculate the slope \( m \)

The formula for \( m \) is:

$$ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} $$

First, compute \( (x_i - \bar{x})(y_i - \bar{y}) \) and \( (x_i - \bar{x})^2 \) for each data point:

\( x_i \)\( y_i \)\( x_i - \bar{x} \)\( y_i - \bar{y} \)\( (x_i - \bar{x})(y_i - \bar{y}) \)\( (x_i - \bar{x})^2 \)
15521.0-7.5-1.0333\( (-7.5)(-1.0333) \approx 7.75 \)\( (-7.5)^2 = 56.25 \)
16021.7-2.5-0.3333\( (-2.5)(-0.3333) \approx 0.833 \)\( (-2.5)^2 = 6.25 \)
16522.42.50.3667\( (2.5)(0.3667) \approx 0.917 \)\( (2.5)^2 = 6.25 \)
17023.17.51.0667\( (7.5)(1.0667) \approx 8.000 \)\( (7.5)^2 = 56.25 \)
17523.712.51.6667\( (12.5)(1.6667) \approx 20.833 \)\( (12.5)^2 = 156.25 \)
Step 3: Sum the products and squared differences
  • \( \sum (x_i - \bar{x})(y_i - \bar{y}) \approx 21.666 + 7.75 + 0.833 + 0.917 + 8.000 + 20.833 = 60 \)
  • \( \sum (x_i - \bar{x})^2 = 156.25 + 56.25 + 6.25 + 6.25 + 56.25 + 156.25 = 437.5 \)
Step 4: Calculate slope \( m \)
$$ m = \frac{60}{437.5} \approx 0.1371 $$
Step 5: Calculate intercept \( b \)

Using \( \bar{y} = m\bar{x} + b \):

$$ 22.0333 = 0.1371(162.5) + b $$
$$ 22.0333 \approx 22.28 + b $$
$$ b \approx 22.0333 - 22.28 \approx -0.2467 $$

Thus, the line of best fit is approximately \( \boldsymbol{y = 0.137x - 0.247} \) (or rounded for simplicity, \( y \approx 0.137x - 0.25 \)).

Part b) Estimate Weight for BMI = 25

Using the line of best fit \( y = 0.137x - 0.247 \) (where \( y = 25 \)):

Step 1: Solve for \( x \)
$$ 25 = 0.137x - 0.247 $$
$$ 25 + 0.247 = 0.137x $$
$$ 25.247 = 0.137x $$
$$ x = \frac{25.247}{0.137} \approx 184.3 $$

Thus, Sam’s estimated weight is approximately 184 pounds (rounded to a reasonable whole number).

Final Answers

a) Line of best fit: \( \boldsymbol{y \approx 0.137x - 0.25} \) (or more precisely \( y \approx 0.137x - 0.247 \))
b) Estimated weight: \( \boldsymbol{\approx 184} \) pounds