If r = 0.5 , xy = 120 , = 8 , x2 = 90 , then value of n is equal to where xy = (x - )(y - ) , x2 = (x - )2

Option b: 10. Solution: Finding n from given correlation data: Given: - r = 0.5 - xy = (x-)(y-) = 120 - = 8 - x2 = (x-)2 = 90 Karl Pearson's formula: r = xy/( x2 × y2) We know 2 = y2/n , so y2 = n 2 = n × 64 = 64n Substituting: 0.5 = 120/(90 × 64n) 0.5 = 120/(5760n) (5760n) = 120/0.5 = 240 5760n = 2402 = 57600 n = 57600/5760 = 10 Hence, Option B is the correct answer.

If $r = 0.5$, $\\sum xy = 120$, $\\sigma_y = 8$, $\\sum x^2 = 90$, then value of n is equal to _________ where $\\sum xy = \\sum (x - \\bar{x})(y - \\bar{y})$, $\\sum x^2 = \\sum (x - \\bar{x})^2$

The correct answer is Option b: $10$. **Finding $n$ from given correlation data:** Given: - $r = 0.5$ - $\\sum xy = \\sum(x-\\bar{x})(y-\\bar{y}) = 120$ - $\\sigma_y = 8$ - $\\sum x^2 = \\sum(x-\\bar{x})^2 = 90$ **Karl Pearson's formula:** $$r = \\frac{\\sum xy}{\\sqrt{\\sum x^2 \\cdot \\sum y^2}}$$ We know $\\sigma_y^2 = \\frac{\\sum y^2}{n}$, so $\\sum y^2 = n \\sigma_y^2 = n \\times 64 = 64n$ Substituting: $$0.5 = \\frac{120}{\\sqrt{90 \\times 64n}}$$ $$0.5 = \\frac{120}{\\sqrt{5760n}}$$ $$\\sqrt{5760n} = \\frac{120}{0.5} = 240$$ $$5760n = 240^2 = 57600$$ $$n = \\frac{57600}{5760} = 10$$ Hence, **Option B** is the correct answer.

Correlation and RegressionMCQMTP June 24 Series IIIQuestion 3671 of 188

All Questions

If $\displaystyle r = 0.5$ , $\displaystyle \\sum xy = 120$ , $\displaystyle \\sigma_y = 8$ , $\displaystyle \\sum x^2 = 90$ , then value of n is equal to _________ where $\displaystyle \\sum xy = \\sum (x - \\bar{x})(y - \\bar{y})$ , $\displaystyle \\sum x^2 = \\sum (x - \\bar{x})^2$

Options

\displaystyle 5

\displaystyle 10

\displaystyle 15

\displaystyle 20

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

Correct Answer

✅ Option b — $\displaystyle 10$

All Options:

A $\displaystyle 5$
B $\displaystyle 10$
C $\displaystyle 15$
D $\displaystyle 20$

Detailed Solution & Explanation

**Finding

\displaystyle n

from given correlation data:** Given: -

\displaystyle r = 0.5

\displaystyle \\sum xy = \\sum(x-\\bar{x})(y-\\bar{y}) = 120

\displaystyle \\sigma_y = 8

\displaystyle \\sum x^2 = \\sum(x-\\bar{x})^2 = 90

**Karl Pearson's formula:**

r = \\frac{\\sum xy}{\\sqrt{\\sum x^2 \\cdot \\sum y^2}}

We know

\displaystyle \\sigma_y^2 = \\frac{\\sum y^2}{n}

, so

\displaystyle \\sum y^2 = n \\sigma_y^2 = n \\times 64 = 64n

Substituting:

0.5 = \\frac{120}{\\sqrt{90 \\times 64n}}

0.5 = \\frac{120}{\\sqrt{5760n}}

\\sqrt{5760n} = \\frac{120}{0.5} = 240

5760n = 240^2 = 57600

n = \\frac{57600}{5760} = 10

Hence, **Option B** 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.

Related Comparison Tables

Correlation vs Regression: Statistics for CA Foundation

Ready to Master Correlation and Regression?

Practice all 188 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free