### ProblemThe 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### ExplanationTo 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.

Let X be the length of the fish in inches and Y be the weight of the fish in pounds From the given information, it is observed that X = Mean length of the fish = 10 inches σx = Standard deviation of length of the fish = 2 inches Y = Mean weight of the fish = 4 pounds σy = Standard deviation of weight of the fish = 0.6 pounds r = Correlation coefficient between length and weight = 0.9The regression line of weight on length of the fish i.e. Y on X in a slope and point form is given by Y-Y = r σyσx(X-X) ⇒ Y-4 = 0.9×0.62(X-10) ⇒ Y-4 = 0.27(X-10) ⇒ Y-4 = 0.27 X - 2.7 ⇒ Y = 1.3 + 0.27 X The regression line of weight on length of the fish i.e. Y on X is Y ^ = 1.3 + 0.27 X If the length of the fish is 15 inches, the weight of the fish is Y ^ = 1.3 + 0.27 ×15 = 1.3 + 4.05 = 5.35 If the length of the fish is 15 inches, the weight of the fish is 5.35

Math

Statistics

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?

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

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

### 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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