Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| A. | Mean (short-cut method) | I. | \(A+\displaystyle\frac{\sum f_iu_i}{\sum f_i}\times i\), where \(u_i=\displaystyle\frac{x_i-A}{i}\) |
| B. | Median | II. | \(l+\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\times i\) |
| C. | Mean (step-deviation method) | III. | \(l+\displaystyle\frac{\frac{N}{2}-C}{f}\times i\) |
| D. | Mode | IV. | \(A+\displaystyle\frac{\sum f_id_i}{\sum f_i}\), where \(d_i=x_i-A\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-III, C-I, D-II
In the formulas, A is the assumed mean, i the class width, l the lower limit of the class concerned, f its frequency and N the total frequency.
A. Mean (short-cut method) — this matches IV. Each deviation \(d_i=x_i-A\) is taken from an assumed mean A, and the mean is A plus the average deviation \(\frac{\sum f_id_i}{\sum f_i}\).
B. Median — this matches III. It starts from the lower limit l of the median class and moves into the class by the share \(\frac{N}{2}-C\), where C is the cumulative frequency of the class before it.
C. Mean (step-deviation method) — this matches I. The deviations are also divided by the class width, \(u_i=\frac{x_i-A}{i}\), so the average of the \(u_i\) has to be multiplied back by i.
D. Mode — this matches II. It uses the frequency \(f_1\) of the modal class and the frequencies \(f_0\) and \(f_2\) of the classes just before and after it.
Tip: The two mean formulas differ only in the “\(\times\,i\)”: the step-deviation formula carries it because each deviation was divided by i.