Measures of Central Tendency and DispersionMCQPYQ June 19Question 2861 of 473
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The AM of 15 observation is 9 and the AM of first 9 observation is 11 and then AM of remaining observation is

Options

A11\displaystyle 11
B6\displaystyle 6
C3\displaystyle 3
D9\displaystyle 9
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Correct Answer

✅ Option b — 6\displaystyle 6

All Options:

  • A11\displaystyle 11
  • B6\displaystyle 6
  • C3\displaystyle 3
  • D9\displaystyle 9

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Detailed Solution & Explanation

**Step 1: Find the total sum of all 15 observations.** Total sum=15×9=135\text{Total sum} = 15 \times 9 = 135 **Step 2: Find the sum of first 9 observations.** Sum of first 9=9×11=99\text{Sum of first 9} = 9 \times 11 = 99 **Step 3: Find the sum of remaining 6 observations.** Remaining observations =15−9=6\displaystyle = 15 - 9 = 6 Sum of remaining 6=135−99=36\text{Sum of remaining 6} = 135 - 99 = 36 **Step 4: Compute the AM of remaining observations.** AM of remaining=366=6\text{AM of remaining} = \frac{36}{6} = 6 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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