Central Tendency & DispersionPYQ Jan. 21Question 3003 of 473
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If there are two groups with n1\displaystyle n_1 and n2\displaystyle n_2 observations and H1\displaystyle H_1 and H2\displaystyle H_2 are respective harmonic means, then the harmonic mean of combined observation is

Options

An1H1+n2H2n1+n2\displaystyle \frac{n_1 H_1 + n_2 H_2}{n_1 + n_2}
Bn1H1+n2H2H1+H2\displaystyle \frac{n_1 H_1 + n_2 H_2}{H_1 + H_2}
Cn1+n2n1H1+n2H2\displaystyle \frac{n_1 + n_2}{\frac{n_1}{H_1} + \frac{n_2}{H_2}}
D(n1+n2)H1H2n1H1+n2H2\displaystyle \frac{(n_1 + n_2)H_1 H_2}{n_1 H_1 + n_2 H_2}
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Correct Answer

✅ Option c — n1+n2n1H1+n2H2\displaystyle \frac{n_1 + n_2}{\frac{n_1}{H_1} + \frac{n_2}{H_2}}

All Options:

  • An1H1+n2H2n1+n2\displaystyle \frac{n_1 H_1 + n_2 H_2}{n_1 + n_2}
  • Bn1H1+n2H2H1+H2\displaystyle \frac{n_1 H_1 + n_2 H_2}{H_1 + H_2}
  • Cn1+n2n1H1+n2H2\displaystyle \frac{n_1 + n_2}{\frac{n_1}{H_1} + \frac{n_2}{H_2}}
  • D(n1+n2)H1H2n1H1+n2H2\displaystyle \frac{(n_1 + n_2)H_1 H_2}{n_1 H_1 + n_2 H_2}

Detailed Solution & Explanation

**Step 1: Recall the formula for Harmonic Mean of a group.** If a group has n\displaystyle n observations with HM = H\displaystyle H, then: ∑1xi=nH\sum \frac{1}{x_i} = \frac{n}{H} **Step 2: For the combined group.** Total sum of reciprocals: ∑all1xi=n1H1+n2H2\sum_{\text{all}} \frac{1}{x_i} = \frac{n_1}{H_1} + \frac{n_2}{H_2} Total observations = n1+n2\displaystyle n_1 + n_2 **Step 3: Combined HM.** H=n1+n2n1H1+n2H2H = \frac{n_1 + n_2}{\frac{n_1}{H_1} + \frac{n_2}{H_2}} This matches **Option C**. Note: The given correct option is A, but mathematically the correct formula for combined HM is Option C. Hence, **Option C** is the correct answer.

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