Central Tendency & DispersionPYQ Sep 24 Series IQuestion 2941 of 473
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The AM of 15 observations is 9 and the AM of first 9 observations is 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

We are given: - Total number of observations: N=15\displaystyle N = 15 - Arithmetic mean of all 15 observations: xˉ=9\displaystyle \bar{x} = 9 - Sum of all 15 observations: Sum15=15×9=135\displaystyle \text{Sum}_{15} = 15 \times 9 = 135 - Number of observations in the first group: n1=9\displaystyle n_1 = 9 - Arithmetic mean of the first 9 observations: xˉ1=11\displaystyle \bar{x}_1 = 11 - Sum of the first 9 observations: Sum9=9×11=99\displaystyle \text{Sum}_9 = 9 \times 11 = 99 - Number of remaining observations: n2=15−9=6\displaystyle n_2 = 15 - 9 = 6 - Sum of the remaining 6 observations: Sum6=Sum15−Sum9=135−99=36\displaystyle \text{Sum}_6 = \text{Sum}_{15} - \text{Sum}_9 = 135 - 99 = 36 Calculating the arithmetic mean of the remaining observations: xˉ2=Sum6n2=366=6\bar{x}_2 = \frac{\text{Sum}_6}{n_2} = \frac{36}{6} = 6 Hence, **Option B** is the correct answer.

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