Class 10 Maths · Original EZT practice
Statistics (Class 10 Chapter 13): summary and practice
Statistics moves from raw lists to grouped data: class marks and the direct method for the mean, assumed-mean and step-deviation shortcuts, the modal class and the mode formula, the median class through cumulative frequency, and the empirical relation linking all three. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 22 October 2026:
1. Mean of grouped data
Replace each class by its class mark, then average with frequencies: the direct method. Assumed-mean and step-deviation methods shrink the arithmetic when values are large or evenly spaced.
2. Mode of grouped data
The modal class is the class with the highest frequency. The mode formula l + (f1 - f0)/(2f1 - f0 - f2) x h refines it using the neighbouring frequencies.
3. Median of grouped data
Build cumulative frequencies, find the class where n/2 falls (the median class), then apply l + (n/2 - cf)/f x h. Cumulative frequency curves - ogives - picture the same computation.
4. The empirical relation
3 Median = Mode + 2 Mean - an approximate bridge between the three measures, handy for finding one when the other two are known.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Mean by the direct method
Choose the method: Mean = sum of values / number of values; for grouped data use class marks weighted by frequency.
EZT practice question
Find the mean of 5, 10, 15, 20, 25.
Show the worked solution
- Sum = 5 + 10 + 15 + 20 + 25 = 75.
- Count = 5.
- Mean = 75/5.
Answer: 15.
Check: The values are symmetric around 15, so the mean must sit exactly at the centre - and it does.
Watch for this: With grouped data, divide by the total FREQUENCY, not by the number of classes.
Now try it yourself
Mean of 2, 4, 6, 8, 10?
Check your answer
6.
30/5.
Mode of grouped data
Choose the method: Mode = l + (f1 - f0)/(2f1 - f0 - f2) x h, where f1 is the modal class frequency.
EZT practice question
A distribution has modal class 20-30 with f1 = 12, previous class frequency f0 = 8, next class frequency f2 = 5, and class width h = 10. Find the mode.
Show the worked solution
- Numerator: f1 - f0 = 12 - 8 = 4.
- Denominator: 2f1 - f0 - f2 = 24 - 8 - 5 = 11.
- Mode = 20 + (4/11) x 10 = 20 + 40/11 = 23.64 (about).
Answer: About 23.64.
Check: The mode must lie INSIDE the modal class - 23.64 sits between 20 and 30, closer to the side with the larger neighbour? f2 < f0 pulls it toward the lower side, and 23.64 is below the midpoint 25. Consistent.
Watch for this: Pick the modal class by the highest FREQUENCY, not by the largest class mark - and read f0 and f2 from the correct neighbours.
Now try it yourself
Same formula with l = 30, f1 = 10, f0 = 6, f2 = 6, h = 10?
Check your answer
30 + 40/8 = 35.
Symmetric neighbours put the mode mid-class.
Median of a raw list
Choose the method: Order the values and take the middle one (or the mean of the two middle ones).
EZT practice question
Find the median of 5, 10, 15, 20, 25.
Show the worked solution
- Already in order.
- Five values - the third is the middle.
- Median = 15.
Answer: 15.
Check: Two values lie below 15 and two above - the median splits the list evenly, as it must.
Watch for this: Always ORDER the data first - the middle entry of an unordered list is not the median.
Now try it yourself
Median of 3, 7, 9, 12, 15?
Check your answer
9.
The third of five ordered values.
Median of grouped data
Choose the method: Median = l + (n/2 - cf)/f x h inside the median class.
EZT practice question
For n = 50, the median class is 20-30 with cumulative frequency before it cf = 18, class frequency f = 14 and width h = 10. Find the median.
Show the worked solution
- n/2 = 25, which falls in 20-30 since cf reaches 18 before it and 18 + 14 = 32 after.
- Median = 20 + (25 - 18)/14 x 10.
- = 20 + 70/14 = 20 + 5.
Answer: 25.
Check: 25 lies inside the median class 20-30, and exactly halfway - with 7 of the class's 14 items needed to reach n/2, halfway is right.
Watch for this: Use the cumulative frequency BEFORE the median class for cf - using the class's own cumulative total overshoots.
Now try it yourself
n = 40, median class 10-20, cf = 10, f = 12, h = 10 - median?
Check your answer
10 + (20 - 10)/12 x 10 = 10 + 100/12 = 18.33 (about).
n/2 = 20.
Use the empirical relation
Choose the method: 3 Median = Mode + 2 Mean.
EZT practice question
A distribution has mode 10 and mean 15. Estimate the median.
Show the worked solution
- 3 Median = 10 + 2 x 15 = 40.
- Median = 40/3.
- = 13.33 (about).
Answer: About 13.33.
Check: The median should lie between mode and mean, closer to the mean: 10 < 13.33 < 15, and 13.33 is nearer 15. The relation places it exactly so.
Watch for this: It is an EMPIRICAL (approximate) relation - quote the result as an estimate, not an exact value.
Now try it yourself
Mode 20, mean 30 - median?
Check your answer
(20 + 60)/3 = 26.67 (about).
3 Median = Mode + 2 Mean.
Assumed-mean and step-deviation for the mean
Choose the method: Mean = a + h x (mean of coded values u = (x - a)/h).
EZT practice question
For 5, 10, 15, 20, 25 with a = 15 and h = 5, find the mean by step deviation.
Show the worked solution
- Coded u: -2, -1, 0, 1, 2.
- Mean of u = 0/5 = 0.
- Mean = 15 + 5 x 0 = 15.
Answer: 15 - matching the direct method.
Check: Every shortcut must reproduce the direct answer; 15 = 15 confirms the coding.
Watch for this: Add a back AFTER multiplying by h - the un-coding is a + h x u-bar, in that order.
Now try it yourself
Coded values for 2, 4, 6, 8, 10 with a = 6, h = 2?
Check your answer
-2, -1, 0, 1, 2.
(x - 6)/2.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Modal class?
Show answer and reason
The class with the highest frequency.
Mode lives where the data piles up.
2. Median class?
Show answer and reason
The class where the cumulative frequency crosses n/2.
The middle of the data by count.
3. Empirical relation?
Show answer and reason
3 Median = Mode + 2 Mean.
Approximate bridge between the three measures.
4. Class mark?
Show answer and reason
(upper limit + lower limit)/2.
One representative value per class.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Mean of 2, 4, 6, 8, 10.
- Median of 3, 7, 9, 12, 15.
- Mode with l = 30, f1 = 10, f0 = 6, f2 = 6, h = 10.
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Questions about this practice page
- When does the step-deviation method save work? (Class 10 Statistics)
- When class marks are large and equally spaced. Coding u = (x - a)/h turns marks like 45, 55, 65 into -1, 0, 1, so the frequency sums stay small; then Mean = a + h x u-bar un-codes the result.
- How are mean, median and mode linked?
- Approximately, by the empirical relation 3 Median = Mode + 2 Mean. Given any two, you can estimate the third - remember it is an approximation, not an identity.
- How do I find the median class?
- Build the cumulative frequency column and halve the total frequency. The first class whose cumulative frequency reaches or passes n/2 is the median class; then apply l + (n/2 - cf)/f x h with cf the cumulative frequency BEFORE that class.
- Are ogives (cumulative frequency curves) in the current book?
- Yes - they appear within the median work in the current NCERT Class 10 Maths book (Reprint 2026-27), as a graphical view of cumulative frequency. There is no separate ogive section; they are part of the median discussion.
- Are these the NCERT exercise questions for Statistics (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.