Measures of Central Tendency and DispersionMCQPYQ June 24 Series IIQuestion 2935 of 473
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The mean of the values of 1,2,3,,n\displaystyle 1, 2, 3, \dots, n with respective frequencies x,2x,3x,,nx\displaystyle x, 2x, 3x, \dots, nx is

Options

An+12\displaystyle \frac{n+1}{2}
Bn\displaystyle n
C2n+13\displaystyle \frac{2n+1}{3}
D2n+12\displaystyle \frac{2n+1}{2}
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Correct Answer

Option c2n+13\displaystyle \frac{2n+1}{3}

All Options:

  • An+12\displaystyle \frac{n+1}{2}
  • Bn\displaystyle n
  • C2n+13\displaystyle \frac{2n+1}{3}
  • D2n+12\displaystyle \frac{2n+1}{2}

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Detailed Solution & Explanation

We are given the values xi=i\displaystyle x_i = i for i=1,2,,n\displaystyle i = 1, 2, \dots, n with respective frequencies fi=ix\displaystyle f_i = ix. 1. Find the sum of the frequencies (total number of observations N\displaystyle N): N=i=1nfi=i=1nix=xi=1ni=xn(n+1)2N = \sum_{i=1}^n f_i = \sum_{i=1}^n ix = x \sum_{i=1}^n i = x \frac{n(n+1)}{2} 2. Find the sum of the products of values and frequencies (fixi\displaystyle \sum f_i x_i): i=1nfixi=i=1n(ix)i=xi=1ni2=xn(n+1)(2n+1)6\sum_{i=1}^n f_i x_i = \sum_{i=1}^n (ix) \cdot i = x \sum_{i=1}^n i^2 = x \frac{n(n+1)(2n+1)}{6} 3. Calculate the mean (xˉ\displaystyle \bar{x}): xˉ=fixiN=xn(n+1)(2n+1)6xn(n+1)2=2n+13\bar{x} = \frac{\sum f_i x_i}{N} = \frac{x \frac{n(n+1)(2n+1)}{6}}{x \frac{n(n+1)}{2}} = \frac{2n+1}{3} Hence, **Option C** is the correct answer.

About This Chapter: Measures of Central Tendency and Dispersion

Paper

Paper 3: Quantitative Aptitude

Weightage

12-15 Marks

Key Topics

Mean, Median, Mode, Range, Mean Deviation, Standard Deviation

The core foundation of Statistics. This chapter covers Mean (Arithmetic, Geometric, Harmonic), Median, Mode, and their properties. It also explores measures of spread like Range, Mean Deviation, Standard Deviation, and Quartile Deviation.

View Official ICAI Syllabus

Exam Strategy Tip

Do not just memorize formulas; ICAI loves asking about the mathematical properties (e.g., 'sum of deviations from the AM is always zero'). You can usually eliminate 2 options just by knowing the properties.

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