Measures of Central Tendency and DispersionMCQMTP Dec 22 Series IIQuestion 2912 of 473
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The AM of 15\displaystyle 15 observations is 9\displaystyle 9 and the AM of first 9\displaystyle 9 observations is 11\displaystyle 11 and then AM of remaining observations is:

Options

A11
B6
C5
D9
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Correct Answer

✅ Option b — 6

All Options:

  • A11
  • B6
  • C5
  • D9

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Detailed Solution & Explanation

**Step 1: Total sum of all 15 observations.** Total=15×9=135\text{Total} = 15 \times 9 = 135 **Step 2: Sum of first 9 observations.** Sum of first 9=9×11=99\text{Sum of first 9} = 9 \times 11 = 99 **Step 3: Sum and mean of remaining 6 observations.** Sum of remaining=135−99=36\text{Sum of remaining} = 135 - 99 = 36 AM of remaining=3615−9=366=6\text{AM of remaining} = \frac{36}{15 - 9} = \frac{36}{6} = 6 **Note:** The correct AM of remaining observations is **6** (Option B), not 11 (Option A). Hence, **Option B** 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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