Statistics Class 10 Notes: Mean, Median, Mode and Ogives
These statistics class 10 notes cover Chapter 13 for the current CBSE session. You will revise how to find the mean, median, and mode for grouped frequency data, understand the class-mark assumption, master three mean methods, and spot the discontinuous-class trap that costs students marks.
From Class 9 Basics to Grouped Data: Why This Chapter Exists
In Class 9, you found the mean, median, and mode of ungrouped data, where each observation was a single number like 28 or 40. You also drew bar graphs, histograms, and frequency polygons (NCERT, p. 1). That method is exact because you know every individual value.
In real-life situations, data sets are large, so we condense them into grouped frequency distributions like 10-25, 25-40, and 40-55. You no longer see individual marks; you only see how many students fell in each interval. The chapter extends the three measures of central tendency to this grouped form.
To compute the mean, you need a single representative value for each class. The textbook assumes the frequency of each class interval is centred at its mid-point, called the class mark (NCERT, p. 1-2). For example, the class mark of 25-40 is the average of its limits: 32.5.
\[ \text{Class mark} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2} \]
This mid-point assumption means every grouped-data answer is an approximation. The ungrouped mean is exact, but the grouped mean is slightly off because it treats all observations in a class as if they sit at the centre. When the grounding compares the exact mean 59.3 to the grouped answer 62 for the same marks data, the difference comes from this assumption (NCERT, p. 2).
Mastering Mean of Grouped Data: Direct, Assumed Mean, and Step-Deviation
There are three methods to find the mean of grouped data. All three give the same answer because they are algebraic rearrangements of one formula. You pick the method based on how large the class marks and frequencies are (NCERT, p. 9).
The Direct Method is simplest. You multiply each class mark by its frequency, add the products, and divide by the total frequency (NCERT, p. 2).
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
When the class marks and frequencies are large, multiplying them directly becomes tedious. The Assumed Mean Method subtracts a chosen value a from each class mark so you work with smaller deviations d_i (NCERT, p. 4-6). It assumes the frequency of each class is centred at the mid-point x_i, so the sum of deviations is simply added back to a.
\[ \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]
If the deviations d_i share a common factor (the class size h), the Step-Deviation Method divides them by h to shrink the numbers further (NCERT, p. 6-9). The formula multiplies the mean of the u_i values back by h and adds a.
\[ \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \]
If class sizes are unequal but the d_i values have a common divisor, you can still use step-deviation by picking h as that divisor (NCERT, p. 9). The value of the mean does not depend on the choice of a or h; the algebra guarantees the result matches the direct method.
| Method | When to use | Key formula |
|---|---|---|
| Direct | Class marks and frequencies are small | \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \) |
| Assumed Mean | Class marks and frequencies are large | \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \) |
| Step-Deviation | Deviations have a common factor h | \( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \) |
For more CBSE Class 10 Mathematics revision pages, see the Class 10 Mathematics notes index. You can also browse the full CBSE notes hub and the Class 10 notes collection.
Definitions Box: Key Statistical Terms in Chapter 13
Every formula in this chapter uses symbols that you must identify correctly before substituting. The table below collects the terms you will meet, their meanings, and a simple original example (NCERT, p. 1, 13, 18, 19).
| Term | Meaning | Example |
|---|---|---|
| Class mark | Mid-point of a class interval; the representative value. | For 30-40, the class mark is 35. |
| Assumed mean (a) | A chosen class mark, usually near the centre of the data, used to shrink numbers. | If class marks are 15, 25, 35, 45, you may take a = 25. |
| Deviation (\( d_i \)) | Difference between each class mark and the assumed mean: \( d_i = x_i – a \). | If a = 25 and \( x_i = 35 \), then \( d_i = 10 \). |
| Class size (h) | Width of each class interval, used to divide deviations in step-deviation. | For 30-40, h = 10. |
| Modal class | The class with the highest frequency. | In 0-10 (4), 10-20 (9), 20-30 (6), it is 10-20. |
| Median class | The class whose cumulative frequency is just greater than \( \frac{n}{2} \). | If n = 30, \( \frac{n}{2} = 15 \); choose the class whose cf first crosses 15. |
| Cumulative frequency (cf) | Running total of frequencies up to a given class. | For 5, 8, 10: cf is 5, 13, 23. |
Formulas and Symbol Meanings for Grouped Statistics
Here are the five formulas you must memorise, with every symbol defined (NCERT, p. 1, 6, 13, 19). The empirical relationship lets you estimate the third measure if you know two (NCERT, p. 27).
- Direct Mean: \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \), where \( f_i \) is frequency and \( x_i \) is the class mark.
- Assumed Mean: \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \), where \( a \) is the assumed mean and \( d_i = x_i – a \).
- Step-Deviation: \( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \), where \( u_i = \frac{x_i – a}{h} \) and \( h \) is the class size.
- Mode: \( \text{Mode} = l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h \), where \( l \) is the lower limit of the modal class, \( f_1 \) is the frequency of the modal class, \( f_0 \) is the frequency of the class preceding the modal class, and \( f_2 \) is the frequency of the class succeeding the modal class.
- Median: \( \text{Median} = l + \left( \frac{\frac{n}{2} – cf}{f} \right) \times h \), where \( l \) is the lower limit of the median class, \( n \) is the total number of observations, \( cf \) is the cumulative frequency of the class preceding the median class, \( f \) is the frequency of the median class, and \( h \) is the class size.
- Empirical relationship: \( 3 \, \text{Median} = \text{Mode} + 2 \, \text{Mean} \). Use this to estimate one measure when you have the other two; it is an approximation, not an identity.
The class-mark and continuity assumptions underlie all three mean formulas because every deviation calculation treats the class mid-point as the representative value. To see how this applies to another geometry-based chapter, compare with the Surface Areas and Volumes notes; for a data-driven extension, read the Probability notes.
Worked Example: Step-Deviation Method for Mean Pocket Money
This worked example uses fresh numbers so you can practise the full column setup from scratch.
Problem: The daily pocket money (in ₹) of 40 students is grouped as: 0-10 (4), 10-20 (8), 20-30 (10), 30-40 (12), 40-50 (6). Find the mean using the step-deviation method.
Step 1: Build the table. Class marks \( x_i \) are mid-points: 5, 15, 25, 35, 45. Choose \( a = 25 \) (central mark) and \( h = 10 \) (class size). Compute \( d_i = x_i – a \) and \( u_i = \frac{x_i – a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( d_i = x_i – 25 \) | \( u_i = \frac{d_i}{10} \) | \( f_i u_i \) |
|---|---|---|---|---|---|
| 0-10 | 4 | 5 | -20 | -2 | -8 |
| 10-20 | 8 | 15 | -10 | -1 | -8 |
| 20-30 | 10 | 25 | 0 | 0 | 0 |
| 30-40 | 12 | 35 | 10 | 1 | 12 |
| 40-50 | 6 | 45 | 20 | 2 | 12 |
| Total | \( \sum f_i = 40 \) | \( \sum f_i u_i = 8 \) |
Step 2: Apply the step-deviation formula.
\[ \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \]
\[ \bar{x} = 25 + \left( \frac{8}{40} \right) \times 10 \]
\[ \bar{x} = 25 + (0.2) \times 10 = 25 + 2 = 27 \]
Final answer: The mean daily pocket money is ₹ 27.
Worked Example: Finding Mode and Identifying the Modal Class
Problem: The number of books read by 50 students in a month: 0-5 (3), 5-10 (7), 10-15 (12), 15-20 (15), 20-25 (8), 25-30 (5). Find the mode.
Step 1: Identify the modal class. The highest frequency is 15, in the class 15-20. So the modal class is 15-20.
Step 2: Read off the symbols. \( l = 15 \), \( h = 5 \), \( f_1 = 15 \), \( f_0 = 12 \) (preceding class), \( f_2 = 8 \) (succeeding class).
Step 3: Substitute into the mode formula.
\[ \text{Mode} = l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h \]
\[ \text{Mode} = 15 + \left( \frac{15 – 12}{2(15) – 12 – 8} \right) \times 5 \]
\[ \text{Mode} = 15 + \left( \frac{3}{30 – 12 – 8} \right) \times 5 = 15 + \left( \frac{3}{10} \right) \times 5 \]
\[ \text{Mode} = 15 + \frac{15}{10} = 15 + 1.5 = 16.5 \]
Final answer: The mode is 16.5 books.
The mode shifts inside the modal class toward the side of the larger neighbour: since \( f_0 = 12 \) is closer to \( f_1 = 15 \) than \( f_2 = 8 \) is, the mode sits nearer the lower limit.
Worked Example: Cumulative Frequency and the Median Formula
Problem: Daily water consumption (in litres) of 60 households: 50-60 (5), 60-70 (12), 70-80 (18), 80-90 (14), 90-100 (11). Find the median.
Step 1: Build the cumulative frequency column.
| Class | \( f_i \) | Cumulative frequency (cf) |
|---|---|---|
| 50-60 | 5 | 5 |
| 60-70 | 12 | 17 |
| 70-80 | 18 | 35 |
| 80-90 | 14 | 49 |
| 90-100 | 11 | 60 |
Step 2: Compute \( \frac{n}{2} \). Here \( n = 60 \), so \( \frac{n}{2} = 30 \). The cumulative frequency just greater than 30 is 35, which belongs to 70-80. So the median class is 70-80.
Step 3: Read the values. \( l = 70 \), \( cf = 17 \) (cumulative frequency of the class preceding the median class), \( f = 18 \), \( h = 10 \).
Step 4: Substitute into the median formula.
\[ \text{Median} = l + \left( \frac{\frac{n}{2} – cf}{f} \right) \times h \]
\[ \text{Median} = 70 + \left( \frac{30 – 17}{18} \right) \times 10 \]
\[ \text{Median} = 70 + \left( \frac{13}{18} \right) \times 10 \]
\[ \text{Median} = 70 + \frac{130}{18} = 70 + 7.22 = 77.22 \]
Final answer: The median water consumption is 77.22 litres. About 50% of households consume less than this, and 50% consume more.
The ‘Continuous Classes’ Trap: Converting Discontinuous Intervals
The median and mode formulas assume continuous classes, meaning there is no gap between the upper limit of one class and the lower limit of the next (NCERT, p. 28, 30). For data like 127-135, 136-144, 145-153, a gap of 1 exists between successive intervals. The formula will not work directly.
The fix is to subtract 0.5 from each lower limit and add 0.5 to each upper limit. This closes the gap by distributing the difference equally on both sides.
| Original class | Adjusted continuous class |
|---|---|
| 118-126 | 117.5-126.5 |
| 127-135 | 126.5-135.5 |
| 136-144 | 135.5-144.5 |
This adjustment matters because the \( l \) you substitute into the median and mode formula must be a true lower limit with no gap to the class below it. The same condition applies when you draw an ogive (NCERT, p. 30).
Common Mistakes Students Make in Statistics Calculations
The five slip-ups below are the most frequent reasons students lose marks. Each has a strict correction and a quick way to check your answer (NCERT, p. 3, 6, 13, 19, 30).
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using the upper limit as the class mark. | Class mark is the mid-point: \( \frac{\text{upper} + \text{lower}}{2} \). | For 30-40, class mark is 35, not 30 or 40. |
| Forgetting to multiply by h in step-deviation. | Always multiply \( \frac{\sum f_i u_i}{\sum f_i} \) by \( h \) and add \( a \). | Units must match the data (₹, kg, mm); if the answer is too small, you missed h. |
| Taking the cf of the median class instead of the preceding class. | In the median formula, cf is the cumulative frequency of the class before the median class. | The cf you use must be less than \( \frac{n}{2} \); if it is greater, you picked the wrong one. |
| Not converting discontinuous classes. | Subtract 0.5 from lower limits and add 0.5 to upper limits before using the formula. | After adjustment, the upper limit of one class must equal the lower limit of the next. |
| Swapping \( f_0 \) and \( f_2 \) in the mode formula. | \( f_0 \) is the frequency of the class preceding the modal class; \( f_2 \) is the class succeeding it. | |
| Count downward and upward from the modal class; write the row order before substituting. |
Exam Notes: What Examiners Look For in Statistics
CBSE marking schemes reward method marks, so write the formula before substitution. A correct numerical answer without the formula can lose marks if the substitution step is unclear (NCERT, p. 11, 16, 28, 30).
Exercise 13.3 question 1 asks you to compute mean, median, and mode for the same data and then interpret them. The comparison step earns a separate mark, so state which measure is larger and what it tells you about the data.
For ogive (cumulative frequency curve) questions, the scale need not be the same on both axes (NCERT, p. 30). This is a deliberate allowance; choose a convenient scale for the class limits on the x-axis and a separate scale for cumulative frequency on the y-axis.
When a question gives a missing frequency \( f \) and a known mean, set up the mean equation and solve for \( f \); remember that all frequencies must sum to the given total.
For Class 9 revision context, you can verify the ungrouped definitions against the same site’s earlier chapter pages.
Quick Revision Summary of Statistics Class 10
Use this compact recap the night before the exam. Each formula assumes continuous classes and the mid-point (class-mark) assumption.
| Measure | Formula | Key symbols |
|---|---|---|
| Direct Mean | \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \) | Use when class marks and frequencies are small. |
| Assumed Mean | \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \) | \( a \) is a central class mark; \( d_i = x_i – a \). |
| Step-Deviation | \( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \) | \( u_i = \frac{x_i – a}{h} \); \( h \) is the class size. |
| Mode | \( l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h \) | \( l \) is the lower limit of the modal class; \( f_0 \) is preceding, \( f_2 \) is succeeding. |
| Median | \( l + \left( \frac{\frac{n}{2} – cf}{f} \right) \times h \) | cf is the cumulative frequency before the median class. |
- Empirical relationship: \( 3 \, \text{Median} = \text{Mode} + 2 \, \text{Mean} \) (NCERT, p. 27). Use this to estimate the third measure when two are known.
- Use mean when all observations matter and no extreme values dominate. Use median when extreme values are present, such as income or wage data. Use mode when you need the most frequent or popular value, such as the most watched TV programme.
Frequently Asked Questions on Statistics Notes
Which method of finding the mean of grouped data should I use in the board exam?
Choose the method that fits the size of your numbers. If the class marks and frequencies are small, the direct method saves time because you avoid the extra \( a \) and \( h \) columns. If the class marks and frequencies are large, use the assumed mean method or the step-deviation method to shrink the numbers before multiplying. The step-deviation method is best when all the deviations share a common factor equal to the class size (NCERT, p. 9). All three give the same answer.
What is the empirical relationship between mean, median, and mode?
The relationship is \( 3 \, \text{Median} = \text{Mode} + 2 \, \text{Mean} \) (NCERT, p. 27). It is an approximation, not an identity. Use it to estimate the third measure when you already have two.
How do you identify the median class in a grouped frequency distribution?
First compute \( \frac{n}{2} \), where \( n \) is the total frequency. Then build the cumulative frequency column. The median class is the one whose cumulative frequency is just greater than \( \frac{n}{2} \). In the median formula, the cf you substitute is the cumulative frequency of the class preceding the median class, not the median class itself.
Why do we need to convert discontinuous class intervals before finding the median?
The median and mode formulas assume continuous classes, where the upper limit of one interval equals the lower limit of the next. When a gap exists, such as 118-126 followed by 127-135, the formula does not apply directly. You subtract 0.5 from each lower limit and add 0.5 to each upper limit to close the gap (NCERT, p. 28, 30). The adjusted classes are then continuous.
For the original chapter content, you can refer to the official NCERT Class 10 Mathematics Chapter 13 Statistics PDF on ncert.nic.in.
Reference: NCERT Class 10 Mathematics textbook, chapter Statistics.
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