The fish in my pond have mean length 10 inches with a standard deviation of 2 inches and mean weight 4 pounds with a standard deviation of .6 pounds. The correlation coefficient of length and weight is .9. If the length of a particular randomly selected fish is reported to be 15 inches, then what should we predict for the weight of that fish using simple linear regression?

MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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### Problem

The fish in my pond have a mean length of 10 inches with a standard deviation of 2 inches and a mean weight of 4 pounds with a standard deviation of 0.6 pounds. The correlation coefficient of length and weight is 0.9. If the length of a particular randomly selected fish is reported to be 15 inches, then what should we predict for the weight of that fish using simple linear regression?

### Options

- 5.35
- 2.52
- 6.55
- 2.5
- NONE OF THE OTHERS

### Explanation

To solve this problem, we can use the formula for the equation of the regression line:
\[ y = \bar{y} + r \cdot \frac{s_y}{s_x} \cdot (x - \bar{x}) \]

Where:
- \( x \) is the independent variable (fish length),
- \( \bar{x} \) is the mean of \( x \) (10 inches),
- \( s_x \) is the standard deviation of \( x \) (2 inches),
- \( y \) is the dependent variable (fish weight),
- \( \bar{y} \) is the mean of \( y \) (4 pounds),
- \( s_y \) is the standard deviation of \( y \) (0.6 pounds),
- \( r \) is the correlation coefficient (0.9).

Substitute the values into the equation to find the predicted weight of the fish.
Transcribed Image Text:### Problem The fish in my pond have a mean length of 10 inches with a standard deviation of 2 inches and a mean weight of 4 pounds with a standard deviation of 0.6 pounds. The correlation coefficient of length and weight is 0.9. If the length of a particular randomly selected fish is reported to be 15 inches, then what should we predict for the weight of that fish using simple linear regression? ### Options - 5.35 - 2.52 - 6.55 - 2.5 - NONE OF THE OTHERS ### Explanation To solve this problem, we can use the formula for the equation of the regression line: \[ y = \bar{y} + r \cdot \frac{s_y}{s_x} \cdot (x - \bar{x}) \] Where: - \( x \) is the independent variable (fish length), - \( \bar{x} \) is the mean of \( x \) (10 inches), - \( s_x \) is the standard deviation of \( x \) (2 inches), - \( y \) is the dependent variable (fish weight), - \( \bar{y} \) is the mean of \( y \) (4 pounds), - \( s_y \) is the standard deviation of \( y \) (0.6 pounds), - \( r \) is the correlation coefficient (0.9). Substitute the values into the equation to find the predicted weight of the fish.
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