Central Tendency & DispersionMTP 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

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.

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