Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\overline{x}\) (Mean) | (I) | Mode + 2 Mean |
| (B) | Mode | (II) | \(l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h\) |
| (C) | Median | (III) | \(\displaystyle\frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}\) |
| (D) | 3 Median | (IV) | \(l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-IV, C-II, D-I
In the grouped-data formulas, \(l\) is the lower limit of the class concerned, \(h\) the class width and \(f\) a frequency.
(A) \(\overline{x}\) (Mean) — this matches (III) \(\frac{\sum f_i x_i}{\sum f_i}\). The mean of a frequency distribution is the sum of each value times its frequency, divided by the total frequency.
(B) Mode — this matches (IV) \(l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h\). Here \(f_1\) is the frequency of the modal class (the class with the highest frequency), and \(f_0\), \(f_2\) are the frequencies of the classes just before and just after it.
(C) Median — this matches (II) \(l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h\). The median class is the one in which the \(\frac{n}{2}\)th value falls; \(cf\) is the cumulative frequency of the class before it.
(D) 3 Median — this matches (I) Mode + 2 Mean. It is the empirical relation Mode = 3 Median − 2 Mean, rearranged.