We appreciate your visit to begin array c c hline x y text Miles Run text Weight in Pounds hline 10 130 hline 1 200 hline 3 167 hline 5. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
- Calculate the mean of the x values: $\bar{x} = 5.4$.
- Calculate the mean of the y values: $\bar{y} = 155.4$.
- Calculate the slope of the regression line: $b \approx -7.33$.
- Calculate the y-intercept of the regression line: $a \approx 194.97$. The predicted weight for 7 miles is approximately $\boxed{150}$.
### Explanation
1. Understanding the Problem
We are given a table of data relating miles run ($x$) to weight in pounds ($y$). The data points are (10, 130), (1, 200), (3, 167), (5, 145), and (8, 135). The question is a multiple-choice question, and we are asked to find the weight when the miles run is 7. The multiple-choice options are: 103, 1199, 130, and 150 (assuming the last option is a typo). We will find a linear regression model to predict the weight based on the miles run.
2. Calculating the Means
First, we need to calculate the mean of the $x$ values and the mean of the $y$ values.
The mean of the $x$ values is: $$\bar{x} = \frac{10 + 1 + 3 + 5 + 8}{5} = \frac{27}{5} = 5.4$$
The mean of the $y$ values is: $$\bar{y} = \frac{130 + 200 + 167 + 145 + 135}{5} = \frac{777}{5} = 155.4$$
3. Calculating the Slope
Next, we calculate the slope ($b$) of the regression line using the formula:
$$b = \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i - \bar{x})^2}$$
$$b = \frac{(10-5.4)(130-155.4) + (1-5.4)(200-155.4) + (3-5.4)(167-155.4) + (5-5.4)(145-155.4) + (8-5.4)(135-155.4)}{(10-5.4)^2 + (1-5.4)^2 + (3-5.4)^2 + (5-5.4)^2 + (8-5.4)^2}$$
$$b = \frac{(4.6)(-25.4) + (-4.4)(44.6) + (-2.4)(11.6) + (-0.4)(-10.4) + (2.6)(-20.4)}{(4.6)^2 + (-4.4)^2 + (-2.4)^2 + (-0.4)^2 + (2.6)^2}$$
$$b = \frac{-116.84 - 196.24 - 27.84 + 4.16 - 53.04}{21.16 + 19.36 + 5.76 + 0.16 + 6.76}$$
$$b = \frac{-389.8}{53.2} = -7.327067669172931$$
So, $b \approx -7.33$.
4. Calculating the Intercept
Now, we calculate the y-intercept ($a$) of the regression line using the formula:
$$a = \bar{y} - b\bar{x}$$
$$a = 155.4 - (-7.327067669172931)(5.4)$$
$$a = 155.4 + 39.56616541353384 = 194.96616541353384$$
So, $a \approx 194.97$.
5. Predicting the Weight
The linear regression equation is:
$$y = a + bx$$
$$y = 194.96616541353384 - 7.327067669172931x$$
Now, we substitute $x = 7$ into the linear regression equation to predict the weight:
$$y = 194.96616541353384 - 7.327067669172931(7)$$
$$y = 194.96616541353384 - 51.28947368421052 = 143.67669172932332$$
So, the predicted weight is approximately 143.68 pounds.
6. Selecting the Closest Value
Finally, we compare the predicted weight (143.68 pounds) with the multiple-choice options (103, 1199, 130, 150) and select the closest value. The closest value is 150.
7. Final Answer
The predicted weight when 7 miles are run is approximately 143.68 pounds. The closest value from the multiple-choice options is 150 pounds.
### Examples
Linear regression can be used in various real-life scenarios, such as predicting house prices based on size, predicting sales based on advertising spend, or predicting crop yield based on rainfall. In this case, we used linear regression to predict a person's weight based on the number of miles they run. This can be useful for fitness tracking and setting personalized goals.
- Calculate the mean of the y values: $\bar{y} = 155.4$.
- Calculate the slope of the regression line: $b \approx -7.33$.
- Calculate the y-intercept of the regression line: $a \approx 194.97$. The predicted weight for 7 miles is approximately $\boxed{150}$.
### Explanation
1. Understanding the Problem
We are given a table of data relating miles run ($x$) to weight in pounds ($y$). The data points are (10, 130), (1, 200), (3, 167), (5, 145), and (8, 135). The question is a multiple-choice question, and we are asked to find the weight when the miles run is 7. The multiple-choice options are: 103, 1199, 130, and 150 (assuming the last option is a typo). We will find a linear regression model to predict the weight based on the miles run.
2. Calculating the Means
First, we need to calculate the mean of the $x$ values and the mean of the $y$ values.
The mean of the $x$ values is: $$\bar{x} = \frac{10 + 1 + 3 + 5 + 8}{5} = \frac{27}{5} = 5.4$$
The mean of the $y$ values is: $$\bar{y} = \frac{130 + 200 + 167 + 145 + 135}{5} = \frac{777}{5} = 155.4$$
3. Calculating the Slope
Next, we calculate the slope ($b$) of the regression line using the formula:
$$b = \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i - \bar{x})^2}$$
$$b = \frac{(10-5.4)(130-155.4) + (1-5.4)(200-155.4) + (3-5.4)(167-155.4) + (5-5.4)(145-155.4) + (8-5.4)(135-155.4)}{(10-5.4)^2 + (1-5.4)^2 + (3-5.4)^2 + (5-5.4)^2 + (8-5.4)^2}$$
$$b = \frac{(4.6)(-25.4) + (-4.4)(44.6) + (-2.4)(11.6) + (-0.4)(-10.4) + (2.6)(-20.4)}{(4.6)^2 + (-4.4)^2 + (-2.4)^2 + (-0.4)^2 + (2.6)^2}$$
$$b = \frac{-116.84 - 196.24 - 27.84 + 4.16 - 53.04}{21.16 + 19.36 + 5.76 + 0.16 + 6.76}$$
$$b = \frac{-389.8}{53.2} = -7.327067669172931$$
So, $b \approx -7.33$.
4. Calculating the Intercept
Now, we calculate the y-intercept ($a$) of the regression line using the formula:
$$a = \bar{y} - b\bar{x}$$
$$a = 155.4 - (-7.327067669172931)(5.4)$$
$$a = 155.4 + 39.56616541353384 = 194.96616541353384$$
So, $a \approx 194.97$.
5. Predicting the Weight
The linear regression equation is:
$$y = a + bx$$
$$y = 194.96616541353384 - 7.327067669172931x$$
Now, we substitute $x = 7$ into the linear regression equation to predict the weight:
$$y = 194.96616541353384 - 7.327067669172931(7)$$
$$y = 194.96616541353384 - 51.28947368421052 = 143.67669172932332$$
So, the predicted weight is approximately 143.68 pounds.
6. Selecting the Closest Value
Finally, we compare the predicted weight (143.68 pounds) with the multiple-choice options (103, 1199, 130, 150) and select the closest value. The closest value is 150.
7. Final Answer
The predicted weight when 7 miles are run is approximately 143.68 pounds. The closest value from the multiple-choice options is 150 pounds.
### Examples
Linear regression can be used in various real-life scenarios, such as predicting house prices based on size, predicting sales based on advertising spend, or predicting crop yield based on rainfall. In this case, we used linear regression to predict a person's weight based on the number of miles they run. This can be useful for fitness tracking and setting personalized goals.
Thanks for taking the time to read begin array c c hline x y text Miles Run text Weight in Pounds hline 10 130 hline 1 200 hline 3 167 hline 5. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
