Exercise 13.3: Median of Grouped Data
Find the median using cumulative frequency and the formula. You can also draw an ogive (cumulative frequency curve) and read the median from the graph.
Extra Practice Questions
These questions cover the same concepts as Exercise 13.3. Try solving them to build confidence before or after the textbook exercise.
The median of the data 3, 8, 5, 7, 6 (when arranged in order) is:
The median from an ogive can be found at the intersection of 'less than' and 'more than' ogives. At what frequency does this occur?
Find the mode of: | Class | $100-120$ | $120-140$ | $140-160$ | $160-180$ | $180-200$ | | Freq | 12 | 14 | 8 | 6 | 10 |
In a frequency distribution, the sum of $f_i x_i = 420$ and $\sum f_i = 30$. The mean is:
In the step deviation method, if $a = 30$, $h = 10$, $\sum f_i u_i = -12$, and $\sum f_i = 40$, find the mean.
In the median formula, $M = l + \frac{N/2 - cf}{f} \times h$, what does $f$ represent?
The mean of the following distribution is 50. Find the missing frequencies $f_1$ and $f_2$ if total frequency is 120. | Class | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ | | Freq | 17 | $f_1$ | 32 | $f_2$ | 19 |
Find the median from: | Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | | Freq | 5 | 10 | 20 | 15 | 10 | 5 |
Find the mode of: | Class | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ | with frequencies 10, 35, 52, 61, 42.
In an examination, the mean marks of 50 students is 60. Later, two students' marks were misread as 40 and 90 instead of 50 and 60. The corrected mean is:
Stuck on a question?
Paste any question from Exercise 13.3 into our AI Maths Solver and get a step-by-step solution instantly. It works for all NCERT questions.
Try AI Solver — FreeCommon Mistakes to Avoid
- ✗Building the cumulative frequency table incorrectly
- ✗Finding N/2 but then identifying the wrong median class
- ✗Errors in applying the median formula
Other Exercises in Chapter 13
Frequently Asked Questions
What is the median formula for grouped data?
Median = l + [(N/2 - cf) / f] × h, where l = lower limit of median class, N = total frequency, cf = cumulative frequency before median class, f = frequency of median class, h = class width.
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