Consider the two regression lines 3x + 2y = 26 & 6x + y = 31 . Find the mean values of x and y .

Option a: x = 4 and y = 7. Solution: The intersection point of the two regression lines gives the mean values of the variables, and . Therefore, we solve the system of equations simultaneously:1) 3x + 2y = 26 2) 6x + y = 31 y = 31 - 6x the expression for y from equation (2) into equation (1): 3x + 2(31 - 6x) = 26 3x + 62 - 12x = 26 -9x = -36 x = 4 x = 4 back into equation (2) to find y : y = 31 - 6(4) = 31 - 24 = 7 , the mean values are = 4 and = 7 , which corresponds to Option A. (Note: The source file indicating Option B is a typo)., Option A is the correct answer.

Consider the two regression lines $3x + 2y = 26$ & $6x + y = 31$. Find the mean values of $x$ and $y$.

$\bar{x} = 4$ and $\bar{y} = 7$

$\bar{x} = 5$ and $\bar{y} = 6$

$\bar{x} = 6$ and $\bar{y} = 4$

Consider the two regression lines $3x + 2y = 26$ & $6x + y = 31$. Find the mean values of $x$ and $y$.

The correct answer is Option a: $\bar{x} = 4$ and $\bar{y} = 7$. The intersection point of the two regression lines gives the mean values of the variables, $\\bar{x}$ and $\\bar{y}$. Therefore, we solve the system of equations simultaneously:\n1) $3x + 2y = 26$\n2) $6x + y = 31 \\implies y = 31 - 6x$\n\nSubstitute the expression for $y$ from equation (2) into equation (1):\n$$3x + 2(31 - 6x) = 26$$\n$$3x + 62 - 12x = 26$$\n$$-9x = -36 \\implies x = 4$$\n\nSubstitute $x = 4$ back into equation (2) to find $y$:\n$$y = 31 - 6(4) = 31 - 24 = 7$$\n\nThus, the mean values are $\\bar{x} = 4$ and $\\bar{y} = 7$, which corresponds to Option A. (Note: The source file indicating Option B is a typo).\n\nHence, **Option A** is the correct answer.

Correlation and RegressionMCQMTP June 22Question 3754 of 188

All Questions

Consider the two regression lines $\displaystyle 3x + 2y = 26$ & $\displaystyle 6x + y = 31$ . Find the mean values of $\displaystyle x$ and $\displaystyle y$ .

Options

\displaystyle \bar{x} = 4

and

\displaystyle \bar{y} = 7

\displaystyle \bar{x} = 5

and

\displaystyle \bar{y} = 6

\displaystyle \bar{x} = 6

and

\displaystyle \bar{y} = 4

DNone of these

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle \bar{x} = 4$ and $\displaystyle \bar{y} = 7$

All Options:

A $\displaystyle \bar{x} = 4$ and $\displaystyle \bar{y} = 7$
B $\displaystyle \bar{x} = 5$ and $\displaystyle \bar{y} = 6$
C $\displaystyle \bar{x} = 6$ and $\displaystyle \bar{y} = 4$
DNone of these

Detailed Solution & Explanation

The intersection point of the two regression lines gives the mean values of the variables,

\displaystyle \\bar{x}

and

\displaystyle \\bar{y}

. Therefore, we solve the system of equations simultaneously:\n1)

\displaystyle 3x + 2y = 26

\n2)

\displaystyle 6x + y = 31 \\implies y = 31 - 6x

\n\nSubstitute the expression for

\displaystyle y

from equation (2) into equation (1):\n

3x + 2(31 - 6x) = 26

3x + 62 - 12x = 26

-9x = -36 \\implies x = 4

\n\nSubstitute

\displaystyle x = 4

back into equation (2) to find

\displaystyle y

:\n

y = 31 - 6(4) = 31 - 24 = 7

\n\nThus, the mean values are

\displaystyle \\bar{x} = 4

and

\displaystyle \\bar{y} = 7

, which corresponds to Option A. (Note: The source file indicating Option B is a typo).\n\nHence, **Option A** is the correct answer.

About This Chapter: Correlation and Regression

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Correlation Coefficient, Regression Equations

This chapter covers Correlation Coefficient, Regression Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Regression

A statistical method used for estimating the relationships between a dependent variable and one or more independent variables.

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Correlation vs Regression: Statistics for CA Foundation

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Consider the two regression lines 3x+2y=26\displaystyle 3x + 2y = 263x+2y=26 & 6x+y=31\displaystyle 6x + y = 316x+y=31. Find the mean values of x\displaystyle xx and y\displaystyle yy.