Statistics extends the mean, median and mode from ungrouped data to grouped data, where observations sit in class intervals. You represent each class by its class mark, find the mean by the direct, assumed mean or step-deviation method, locate the modal class and apply the mode formula, and build cumulative frequency tables to find the median class and the median. The chapter ends with choosing the right measure for a situation and the empirical relation 3 Median = Mode + 2 Mean. Board questions are table-driven calculations, so a neat, complete table is where the marks come from.
Key Concepts
1. Grouped Data and the Class Mark
Real surveys are large, so the data is condensed into class intervals such as 10 - 25, 25 - 40, 40 - 55. This is grouped data. The individual observations inside a class are then unknown.
To do any calculation, each class needs one value to stand for all the observations in it. NCERT makes one assumption: the frequency of each class is centred around its mid-point. This mid-point is called the class mark.
Class mark = (Upper class limit + Lower class limit) / 2
- Class 10 - 25: class mark = (10 + 25)/2 = 17.5
- Class 150 - 250: class mark = (150 + 250)/2 = 200
Boundary convention: an observation equal to an upper class limit is counted in the next class. In NCERT’s marks example, the 4 students who scored exactly 40 go in 40 - 55, not in 25 - 40. So in a class written as 25 - 40, the value 25 is included and 40 is excluded.
Class size (h) is the difference between the upper and lower limits of a class. For 10 - 25 it is 15.
2. Mean of Grouped Data: the Direct Method
The mean is the sum of all observations divided by the number of observations. If x1, x2, …, xn occur with frequencies f1, f2, …, fn, then the sum of the observations is f1x1 + f2x2 + … + fnxn and the number of observations is f1 + f2 + … + fn. Using Σ (capital sigma, meaning “sum of”):
x̄ = Σfixi / Σfi
For grouped data, xi is the class mark of the ith class. This is the Direct Method.
NCERT illustration. Marks of 30 students grouped with class size 15:
| Class interval | fi | xi | fixi |
|---|---|---|---|
| 10 - 25 | 2 | 17.5 | 35.0 |
| 25 - 40 | 3 | 32.5 | 97.5 |
| 40 - 55 | 7 | 47.5 | 332.5 |
| 55 - 70 | 6 | 62.5 | 375.0 |
| 70 - 85 | 6 | 77.5 | 465.0 |
| 85 - 100 | 6 | 92.5 | 555.0 |
| Total | 30 | 1860.0 |
x̄ = 1860/30 = 62.
Why the answer differs from the ungrouped mean. The same 30 marks, taken one by one, give a mean of 1779/30 = 59.3. The grouped mean is 62. The difference comes from the mid-point assumption: 59.3 is the exact mean and 62 is an approximate mean.
3. Assumed Mean and Step-deviation Methods
When xi and fi are large, the products fixi become tedious. The frequencies cannot be changed, but each xi can be made smaller by subtracting a fixed number from it.
Assumed Mean Method.
- Choose one class mark as the assumed mean a. A class mark near the centre of the list keeps the numbers small.
- Find the deviation di = xi − a for every class.
- Find fidi and add them to get Σfidi.
- Use x̄ = a + Σfidi / Σfi.
Why it works. The mean of the deviations is d̄ = Σfi(xi − a)/Σfi = Σfixi/Σfi − aΣfi/Σfi = x̄ − a. So x̄ = a + d̄.
For the marks data with a = 47.5: di = −30, −15, 0, 15, 30, 45 and fidi = −60, −45, 0, 90, 180, 270. Σfidi = 435, so x̄ = 47.5 + 435/30 = 47.5 + 14.5 = 62, the same as the direct method.
The value of the mean does not depend on the choice of a. NCERT Activity 1 asks you to take each class mark 17.5, 32.5, … as a in turn; every choice gives 62.
Step-deviation Method. In the table above every di is a multiple of 15, the class size. Divide each deviation by h:
ui = (xi − a) / h, and then x̄ = a + h × (Σfiui / Σfi)
This follows because ū = (x̄ − a)/h, so x̄ = a + hū. For the marks data, ui = −2, −1, 0, 1, 2, 3 and fiui = −4, −3, 0, 6, 12, 18. Σfiui = 29, so x̄ = 47.5 + 15 × 29/30 = 47.5 + 14.5 = 62.
Choosing a method (NCERT). All three methods give the same mean, and the two shortcuts are simplified forms of the direct method. Use the direct method when xi and fi are small, and the assumed mean or step-deviation method when they are large. Step-deviation is convenient when all the di have a common factor. With unequal class sizes, take h as a suitable divisor of all the di.
NCERT Example 3 shows the last point. Wickets taken by 45 bowlers are grouped in unequal classes. With a = 200 and h = 20, the class mark 125 gives ui = −75/20 = −3.75, a decimal, which is allowed. Σfiui = −106 and the mean is 200 + 20 × (−106/45) = 200 − 47.11 = 152.89 wickets.
4. Mode of Grouped Data
The mode is the observation that occurs most often, that is, the value with the maximum frequency. For ungrouped data you read it straight from the frequency table. For example, a bowler’s wickets in 10 matches are 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. The value 2 occurs 3 times, more than any other, so the mode is 2.
If more than one value has the same maximum frequency, the data is multimodal. The chapter deals only with problems that have a single mode.
For grouped data you cannot see the most frequent value, only the class with the maximum frequency. This class is the modal class. The mode is a value inside it, found by:
Mode = l + [(f1 − f0) / (2f1 − f0 − f2)] × h
Here l = lower limit of the modal class, h = class size (all classes equal), f1 = frequency of the modal class, f0 = frequency of the class preceding it, and f2 = frequency of the class succeeding it.
NCERT Example 5. Family sizes of 20 households: 1 - 3 (7), 3 - 5 (8), 5 - 7 (2), 7 - 9 (2), 9 - 11 (1). The highest frequency is 8, so the modal class is 3 - 5, with l = 3, h = 2, f1 = 8, f0 = 7, f2 = 2.
Mode = 3 + [(8 − 7)/(16 − 7 − 2)] × 2 = 3 + 2/7 = 3.286.
How to read the formula. The mode is pulled towards the neighbouring class with the larger frequency. Here f0 = 7 is much larger than f2 = 2, so the mode sits near the lower end of 3 - 5. When f0 = f2, the fraction becomes 1/2 and the mode is exactly the class mark of the modal class.
Comparing mode and mean (NCERT Example 6). For the 30 students’ marks, the modal class is 40 - 55 with f1 = 7, f0 = 3, f2 = 6. Mode = 40 + [(7 − 3)/(14 − 6 − 3)] × 15 = 40 + (4/5) × 15 = 52. The mean was 62. So the maximum number of students obtained 52 marks, while on average a student obtained 62 marks. The mode may be less than, equal to, or more than the mean depending on the data.
The chapter uses this formula only for equal class sizes.
5. Cumulative Frequency Distributions
The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class to the frequency of that class.
NCERT uses the marks of 53 students (out of 100) in classes 0 - 10, 10 - 20, …, 90 - 100 with frequencies 5, 3, 4, 3, 3, 4, 7, 9, 7, 8.
| Less than type | Cumulative frequency | More than or equal to type | Cumulative frequency |
|---|---|---|---|
| Less than 10 | 5 | ≥ 0 | 53 |
| Less than 20 | 8 | ≥ 10 | 48 |
| Less than 30 | 12 | ≥ 20 | 45 |
| Less than 40 | 15 | ≥ 30 | 41 |
| Less than 50 | 18 | ≥ 40 | 38 |
| Less than 60 | 22 | ≥ 50 | 35 |
| Less than 70 | 29 | ≥ 60 | 31 |
| Less than 80 | 38 | ≥ 70 | 24 |
| Less than 90 | 45 | ≥ 80 | 15 |
| Less than 100 | 53 | ≥ 90 | 8 |
- In the less than type, the numbers 10, 20, …, 100 are the upper limits of the classes. The column keeps growing and ends at n.
- In the more than type, the numbers 0, 10, …, 90 are the lower limits of the classes. The column starts at n and keeps falling.
- A table with only the class column and the cumulative frequency column is a cumulative frequency table.
Going backwards. To recover class frequencies from a cumulative table, subtract consecutive cumulative frequencies. In NCERT Example 7 the heights of 51 girls are given as less than 140 (4), less than 145 (11), less than 150 (29), less than 155 (40), less than 160 (46), less than 165 (51). The frequencies are: below 140: 4; 140 - 145: 11 − 4 = 7; 145 - 150: 29 − 11 = 18; 150 - 155: 40 − 29 = 11; 155 - 160: 46 − 40 = 6; 160 - 165: 51 − 46 = 5.
6. Median of Grouped Data
The median is the middle-most observation. For ungrouped data, arrange the values in ascending order. If n is odd, the median is the ((n + 1)/2)th observation. If n is even, it is the average of the (n/2)th and (n/2 + 1)th observations.
Ungrouped with frequencies (NCERT). Marks of 100 students: 20 (6), 25 (20), 28 (24), 29 (28), 33 (15), 38 (4), 42 (2), 43 (1). The cumulative frequencies are 6, 26, 50, 78, 93, 97, 99, 100. Since n = 100 is even, the median is the average of the 50th and 51st observations. These are 28 and 29, so Median = (28 + 29)/2 = 28.5.
Grouped data. The middle observation now lies somewhere inside a class, so we find the class first and then a value inside it.
- Find n = Σfi and n/2.
- Write the cumulative frequency column.
- The median class is the class whose cumulative frequency is greater than (and nearest to) n/2.
- Apply Median = l + [(n/2 − cf) / f] × h.
Here l = lower limit of the median class, n = number of observations, cf = cumulative frequency of the class preceding the median class, f = frequency of the median class, and h = class size (all classes equal).
NCERT illustration. For the 53 students above, n/2 = 26.5. The cumulative frequency first goes above 26.5 at 29, in the class 60 - 70. So l = 60, cf = 22, f = 7, h = 10.
Median = 60 + [(26.5 − 22)/7] × 10 = 60 + 45/7 = 66.4. About half the students scored less than 66.4 and the other half scored more.
The idea behind the formula. Up to the lower limit l, cf observations are already counted. We need n/2 − cf more observations from inside the median class. The class holds f observations spread evenly over a width h, so the median lies [(n/2 − cf)/f] × h beyond l.
The median of grouped data with unequal class sizes can also be calculated, but the chapter does not discuss it.
7. Continuous Classes
The mode and median formulas assume continuous classes, where the upper limit of one class equals the lower limit of the next. NCERT’s note to the reader states that for calculating mode and median of grouped data, the class intervals must be continuous before the formulae are applied.
Some data comes in classes with gaps, such as 118 - 126, 127 - 135, 136 - 144. The gap is 127 − 126 = 1. Subtract half the gap (0.5) from every lower limit and add 0.5 to every upper limit.
The classes become 117.5 - 126.5, 126.5 - 135.5, …, 171.5 - 180.5, exactly as the NCERT hint for Exercise 13.3 Question 4 says. The class size is now 9, and the class marks are unchanged (122, 131, …). For the mean, only class marks are needed, so conversion makes no difference there. For the mode and median, l and h come from the converted classes.
8. Choosing a Measure and the Empirical Relation
| Measure | Best used when | NCERT examples |
|---|---|---|
| Mean | All observations matter and you want to compare distributions. It uses every observation and lies between the smallest and largest values. | Comparing average results of different schools in one examination |
| Median | Individual observations are not important, you want a “typical” value, and extreme values are present | Typical productivity rate of workers, average wage in a country |
| Mode | You want the most frequent value or the most popular item | Most popular TV programme, consumer item in greatest demand, colour of vehicle used by most people |
Weakness of the mean. Extreme values affect it. NCERT’s example: if one class has frequency 2 and the five others have 20, 25, 20, 21 and 18, the mean will not reflect how the data behaves.
Mean or mode for marks. If you want the average marks obtained by the students, use the mean. If you want the marks obtained by most of the students, use the mode.
Empirical relationship:
3 Median = Mode + 2 Mean
It is an observed (empirical) relation and gives an approximate link, so use it when a question asks for one measure from the other two. Rearranged: Mode = 3 Median − 2 Mean, Mean = (3 Median − Mode)/2, Median = (Mode + 2 Mean)/3.
Example: if the median is 26 and the mean is 27, then Mode = 3 × 26 − 2 × 27 = 78 − 54 = 24.
Formula and Theorem Sheet
| Result | Formula | When to use |
|---|---|---|
| Class mark | (Upper limit + Lower limit)/2 | Every mean calculation on grouped data |
| Direct method | x̄ = Σfixi / Σfi | Small xi and fi |
| Assumed mean method | x̄ = a + Σfidi / Σfi, di = xi − a | Large xi |
| Step-deviation method | x̄ = a + h(Σfiui / Σfi), ui = (xi − a)/h | Large xi with a common factor in all di |
| Mode | l + [(f1 − f0)/(2f1 − f0 − f2)] × h | Modal class = class of highest frequency |
| Median | l + [(n/2 − cf)/f] × h | Median class = first class with cf > n/2 |
| Empirical relation | 3 Median = Mode + 2 Mean | Find one measure from the other two |
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Important Definitions
- Grouped data: data condensed into class intervals, each with a frequency.
- Class mark (mid-point): the average of the upper and lower limits of a class. It represents all observations in that class.
- Mode: the value among the observations that occurs most often, that is, the value with the maximum frequency.
- Modal class: the class with the maximum frequency in a grouped distribution.
- Median: the value of the middle-most observation when data is arranged in order.
- Cumulative frequency: the frequency obtained by adding the frequencies of all the preceding classes to the frequency of the given class.
- Cumulative frequency distribution of the less than type: a table giving the number of observations less than each upper class limit.
- Cumulative frequency distribution of the more than type: a table giving the number of observations greater than or equal to each lower class limit.
- Median class: the class whose cumulative frequency is greater than and nearest to n/2.
Solved Examples (NCERT-Based)
Example 1: Plants in 20 houses (NCERT Exercise 13.1, Q1)
Question: A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Number of plants: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10, 10 - 12, 12 - 14; number of houses: 1, 2, 1, 5, 6, 2, 3. Which method did you use for finding the mean, and why?
Class marks: 1, 3, 5, 7, 9, 11, 13. fixi = 1, 6, 5, 35, 54, 22, 39. Σfi = 20 and Σfixi = 162.
x̄ = 162/20 = 8.1 plants per house. The direct method was used because the class marks and frequencies are small.
Example 2: Daily wages of 50 workers (NCERT Exercise 13.1, Q2)
Question: Consider the following distribution of daily wages of 50 workers of a factory. Daily wages (in ₹) 500 - 520, 520 - 540, 540 - 560, 560 - 580, 580 - 600 with number of workers 12, 14, 8, 6, 10. Find the mean daily wages of the workers of the factory by using an appropriate method.
The class marks are large (510, 530, 550, 570, 590) and all deviations from 550 are multiples of 20. Use step-deviation with a = 550, h = 20.
| xi | fi | ui | fiui |
|---|---|---|---|
| 510 | 12 | −2 | −24 |
| 530 | 14 | −1 | −14 |
| 550 | 8 | 0 | 0 |
| 570 | 6 | 1 | 6 |
| 590 | 10 | 2 | 20 |
| Total | 50 | −12 |
x̄ = 550 + 20 × (−12/50) = 550 − 4.8 = ₹ 545.20.
Example 3: Missing frequency from the mean (NCERT Exercise 13.1, Q3)
Question: The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Classes 11 - 13, 13 - 15, 15 - 17, 17 - 19, 19 - 21, 21 - 23, 23 - 25 with frequencies 7, 6, 9, 13, f, 5, 4.
Class marks: 12, 14, 16, 18, 20, 22, 24. Take a = 18, h = 2, so ui = −3, −2, −1, 0, 1, 2, 3.
fiui: −21, −12, −9, 0, f, 10, 12. Σfiui = f − 20. Σfi = 44 + f.
18 = 18 + 2 × (f − 20)/(44 + f) ⇒ (f − 20)/(44 + f) = 0 ⇒ f = 20.
Check with the direct method: Σfixi = 84 + 84 + 144 + 234 + 400 + 110 + 96 = 1152 and Σfi = 64. 1152/64 = 18. Correct.
Example 4: Mangoes in boxes (NCERT Exercise 13.1, Q5)
Question: In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Number of mangoes 50 - 52, 53 - 55, 56 - 58, 59 - 61, 62 - 64; number of boxes 15, 110, 135, 115, 25. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
The classes are not continuous, but the mean needs only class marks: 51, 54, 57, 60, 63. The frequencies are large, so use step-deviation with a = 57, h = 3: ui = −2, −1, 0, 1, 2.
fiui = −30, −110, 0, 115, 50. Σfiui = 25, Σfi = 400.
x̄ = 57 + 3 × 25/400 = 57 + 0.1875 = 57.19 mangoes (approximately). The step-deviation method was chosen because the frequencies are large and the deviations share the factor 3.
Example 5: Mode and mean of patients’ ages (NCERT Exercise 13.2, Q1)
Question: The following table shows the ages of the patients admitted in a hospital during a year: Ages (in years) 5 - 15, 15 - 25, 25 - 35, 35 - 45, 45 - 55, 55 - 65 with number of patients 6, 11, 21, 23, 14, 5. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Mode: modal class 35 - 45 (frequency 23). l = 35, h = 10, f1 = 23, f0 = 21, f2 = 14.
Mode = 35 + [(23 − 21)/(46 − 21 − 14)] × 10 = 35 + 20/11 = 35 + 1.82 = 36.82 years.
Mean: class marks 10, 20, 30, 40, 50, 60. fixi = 60, 220, 630, 920, 700, 300. Σfixi = 2830, Σfi = 80. x̄ = 2830/80 = 35.38 years.
Interpretation: the largest number of patients admitted were about 36.8 years old, while the average age of a patient was about 35.4 years.
Example 6: Modal lifetime of components (NCERT Exercise 13.2, Q2)
Question: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: lifetimes 0 - 20, 20 - 40, 40 - 60, 60 - 80, 80 - 100, 100 - 120 with frequencies 10, 35, 52, 61, 38, 29. Determine the modal lifetimes of the components.
Modal class 60 - 80. l = 60, h = 20, f1 = 61, f0 = 52, f2 = 38.
Mode = 60 + [(61 − 52)/(122 − 52 − 38)] × 20 = 60 + (9/32) × 20 = 60 + 5.625 = 65.625 hours.
Example 7: Median, mean and mode of electricity use (NCERT Exercise 13.3, Q1)
Question: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Units | fi | cf | xi | ui | fiui |
|---|---|---|---|---|---|
| 65 - 85 | 4 | 4 | 75 | −3 | −12 |
| 85 - 105 | 5 | 9 | 95 | −2 | −10 |
| 105 - 125 | 13 | 22 | 115 | −1 | −13 |
| 125 - 145 | 20 | 42 | 135 | 0 | 0 |
| 145 - 165 | 14 | 56 | 155 | 1 | 14 |
| 165 - 185 | 8 | 64 | 175 | 2 | 16 |
| 185 - 205 | 4 | 68 | 195 | 3 | 12 |
| Total | 68 | 7 |
Median: n/2 = 34. The first cf above 34 is 42, so the median class is 125 - 145 with l = 125, cf = 22, f = 20, h = 20. Median = 125 + [(34 − 22)/20] × 20 = 125 + 12 = 137 units.
Mean: a = 135, h = 20. x̄ = 135 + 20 × 7/68 = 135 + 2.06 = 137.06 units.
Mode: modal class 125 - 145, f1 = 20, f0 = 13, f2 = 14. Mode = 125 + [7/(40 − 13 − 14)] × 20 = 125 + 140/13 = 125 + 10.77 = 135.77 units.
Comparison: all three measures are close to 136 to 137 units.
Example 8: Missing frequencies from the median (NCERT Exercise 13.3, Q2)
Question: If the median of the distribution given below is 28.5, find the values of x and y. Classes 0 - 10 (5), 10 - 20 (x), 20 - 30 (20), 30 - 40 (15), 40 - 50 (y), 50 - 60 (5); total 60.
Total: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15 … (1)
Cumulative frequencies: 5, 5 + x, 25 + x, 40 + x, 40 + x + y, 60.
Median 28.5 lies in 20 - 30, so l = 20, cf = 5 + x, f = 20, h = 10, n/2 = 30.
28.5 = 20 + [(30 − 5 − x)/20] × 10 ⇒ 8.5 = (25 − x)/2 ⇒ 17 = 25 − x ⇒ x = 8.
From (1), y = 7.
Example 9: Median age of policy holders (NCERT Exercise 13.3, Q3)
Question: A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. Below 20 (2), below 25 (6), below 30 (24), below 35 (45), below 40 (78), below 45 (89), below 50 (92), below 55 (98), below 60 (100).
This is a less than type table, so subtract consecutive values to get frequencies: 30 - 35: 45 − 24 = 21; 35 - 40: 78 − 45 = 33.
n = 100, n/2 = 50. The cumulative frequency first exceeds 50 at 78 (below 40), so the median class is 35 - 40. l = 35, cf = 45, f = 33, h = 5.
Median = 35 + [(50 − 45)/33] × 5 = 35 + 25/33 = 35 + 0.76 = 35.76 years.
Example 10: Median length of leaves (NCERT Exercise 13.3, Q4)
Question: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: 118 - 126 (3), 127 - 135 (5), 136 - 144 (9), 145 - 153 (12), 154 - 162 (5), 163 - 171 (4), 172 - 180 (2). Find the median length of the leaves.
Make the classes continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit: 117.5 - 126.5, 126.5 - 135.5, 135.5 - 144.5, 144.5 - 153.5, and so on. Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40.
n/2 = 20. The first cf above 20 is 29, so the median class is 144.5 - 153.5. l = 144.5, cf = 17, f = 12, h = 9.
Median = 144.5 + [(20 − 17)/12] × 9 = 144.5 + 2.25 = 146.75 mm.
Example 11: Weights of 30 students (NCERT Exercise 13.3, Q7)
Question: The distribution below gives the weights of 30 students of a class. Weight (in kg) 40 - 45, 45 - 50, 50 - 55, 55 - 60, 60 - 65, 65 - 70, 70 - 75 with number of students 2, 3, 8, 6, 6, 3, 2. Find the median weight of the students.
Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30. n/2 = 15. The first cf above 15 is 19, so the median class is 55 - 60. l = 55, cf = 13, f = 6, h = 5.
Median = 55 + [(15 − 13)/6] × 5 = 55 + 10/6 = 55 + 1.67 = 56.67 kg.
Competency-Based Questions (with answers)
1. Case-based: Screen time survey
A school surveyed 50 students about the minutes they spend on a phone screen each evening. The results were: 0 - 30 (8), 30 - 60 (12), 60 - 90 (20), 90 - 120 (6), 120 - 150 (4).
(a) Find the modal class and the mode. Modal class 60 - 90. l = 60, h = 30, f1 = 20, f0 = 12, f2 = 6. Mode = 60 + [8/(40 − 12 − 6)] × 30 = 60 + 240/22 = 60 + 10.91 = 70.91 minutes.
(b) Find the median. Cumulative frequencies: 8, 20, 40, 46, 50. n/2 = 25, median class 60 - 90, cf = 20, f = 20. Median = 60 + (5/20) × 30 = 67.5 minutes.
(c) Find the mean by the step-deviation method. a = 75, h = 30, ui = −2, −1, 0, 1, 2; fiui = −16, −12, 0, 6, 8; sum = −14. x̄ = 75 + 30 × (−14/50) = 75 − 8.4 = 66.6 minutes.
2. Case-based: Heights of saplings
A nursery measured 40 saplings and reported: less than 10 cm: 3, less than 20 cm: 10, less than 30 cm: 24, less than 40 cm: 36, less than 50 cm: 40.
(a) Write the frequency of each class. 0 - 10: 3; 10 - 20: 7; 20 - 30: 14; 30 - 40: 12; 40 - 50: 4.
(b) Find the median height. n/2 = 20. The median class is 20 - 30 with cf = 10, f = 14, h = 10. Median = 20 + (10/14) × 10 = 20 + 7.14 = 27.14 cm.
3. Assertion-Reason
Assertion (A): For the marks data in classes 10 - 25, …, 85 - 100, taking a = 17.5 or a = 62.5 in the assumed mean method gives the same mean, 62.
Reason (R): The value of the mean obtained does not depend on the choice of the assumed mean a.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). Since x̄ = a + d̄ and d̄ = x̄ − a, any choice of a gives back the same x̄. R explains A.
4. Assertion-Reason
Assertion (A): The mode of a grouped distribution is always the class mark of the modal class.
Reason (R): The modal class is the class with the maximum frequency.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (d). R is the definition of the modal class. A is false: the mode equals the class mark only when f0 = f2. In NCERT Example 5 the modal class is 3 - 5 (class mark 4) but the mode is 3.286.
5. Assertion-Reason
Assertion (A): If the mode of a distribution is 24 and its mean is 27, then its median is 26.
Reason (R): 3 Median = Mode + 2 Mean.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). 3 Median = 24 + 54 = 78, so Median = 26.
6. Source-based: Choosing a measure
NCERT says: “The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes… However, extreme values in the data affect the mean.”
(a) Give one reason the mean is widely used. It uses every observation and allows two or more distributions to be compared, such as the average results of different schools.
(b) Five classes have frequencies 20, 25, 20, 21, 18 and a sixth has frequency 2. Why might the mean mislead here? One class has a frequency far smaller than the other five, which are more or less equal. NCERT says the mean will certainly not reflect the way such data behaves, so it is a poor representative here.
(c) Name the better measure for the average wage in a country. The median, because wages include extreme values.
7. Error analysis: The wrong cf
A student finds the median of the weights in Example 11 as follows: median class 55 - 60, l = 55, cf = 19, f = 6, h = 5, so Median = 55 + [(15 − 19)/6] × 5 = 51.67 kg. Find the error.
Answer: cf must be the cumulative frequency of the class preceding the median class, which is 13, not 19 (19 belongs to the median class itself). Correct value: 55 + (2/6) × 5 = 56.67 kg.
8. Competency MCQ: Converting classes
Marks in a test are grouped as 1 - 10, 11 - 20, 21 - 30, 31 - 40, 41 - 50 with frequencies 5, 9, 15, 7, 4. For the mode formula, the value of l is:
(a) 21 (b) 20 (c) 20.5 (d) 25.5
Answer: (c) 20.5. The classes have gaps of 1, so they become 0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5, and so on. The modal class is 20.5 - 30.5, so l = 20.5 and h = 10. Mode = 20.5 + [6/(30 − 9 − 7)] × 10 = 20.5 + 4.29 = 24.79.
Important Questions for Board Exams
1-Mark Questions
Q1. Find the class mark of the class 150 - 250.
Answer: (150 + 250)/2 = 200.
Q2. In the step-deviation method, if xi = 62.5, a = 47.5 and h = 15, find ui.
Answer: ui = (62.5 − 47.5)/15 = 1.
Q3. If the mean of a distribution is 40 and its median is 42, find the mode.
Answer: Mode = 3 × 42 − 2 × 40 = 126 − 80 = 46.
Q4. For a less than type cumulative frequency distribution, which limits of the classes are used?
Answer: The upper limits of the class intervals.
Q5. Which measure of central tendency is best for finding the most popular TV programme?
Answer: The mode.
2-Mark Questions
Q6. Find the mean of: classes 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50 with frequencies 4, 6, 10, 6, 4.
Answer: Class marks 5, 15, 25, 35, 45. Σfixi = 20 + 90 + 250 + 210 + 180 = 750. Σfi = 30. Mean = 750/30 = 25.
Q7. Convert to a more than or equal to type table: 0 - 10 (5), 10 - 20 (3), 20 - 30 (4), 30 - 40 (8).
Answer: n = 20. ≥ 0: 20; ≥ 10: 15; ≥ 20: 12; ≥ 30: 8.
3-Mark Questions
Q8. The table below shows the daily expenditure on food of 25 households in a locality: 100 - 150 (4), 150 - 200 (5), 200 - 250 (12), 250 - 300 (2), 300 - 350 (2). Find the mean daily expenditure on food by a suitable method. (NCERT Exercise 13.1, Q6)
Answer: Class marks 125, 175, 225, 275, 325. Take a = 225, h = 50: ui = −2, −1, 0, 1, 2; fiui = −8, −5, 0, 2, 4; sum = −7. Mean = 225 + 50 × (−7/25) = 225 − 14 = ₹ 211.
Q9. The following table gives the distribution of the life time of 400 neon lamps: 1500 - 2000 (14), 2000 - 2500 (56), 2500 - 3000 (60), 3000 - 3500 (86), 3500 - 4000 (74), 4000 - 4500 (62), 4500 - 5000 (48). Find the median life time of a lamp. (NCERT Exercise 13.3, Q5)
Answer: n = 400, n/2 = 200. Cumulative frequencies: 14, 70, 130, 216, … The median class is 3000 - 3500 with cf = 130, f = 86, h = 500. Median = 3000 + (70/86) × 500 = 3000 + 406.98 = 3406.98 hours.
Q10. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches: 3000 - 4000 (4), 4000 - 5000 (18), 5000 - 6000 (9), 6000 - 7000 (7), 7000 - 8000 (6), 8000 - 9000 (3), 9000 - 10000 (1), 10000 - 11000 (1). Find the mode of the data. (NCERT Exercise 13.2, Q5)
Answer: Modal class 4000 - 5000, f1 = 18, f0 = 4, f2 = 9, h = 1000. Mode = 4000 + [14/(36 − 4 − 9)] × 1000 = 4000 + 14000/23 = 4000 + 608.7 = 4608.7 runs.
5-Mark Questions
Q11. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: number of letters 1 - 4 (6), 4 - 7 (30), 7 - 10 (40), 10 - 13 (16), 13 - 16 (4), 16 - 19 (4). Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames. (NCERT Exercise 13.3, Q6)
Answer: Median: cumulative frequencies 6, 36, 76, 92, 96, 100. n/2 = 50, median class 7 - 10, l = 7, cf = 36, f = 40, h = 3. Median = 7 + (14/40) × 3 = 7 + 1.05 = 8.05.
Mean: class marks 2.5, 5.5, 8.5, 11.5, 14.5, 17.5. Take a = 8.5, h = 3: ui = −2, −1, 0, 1, 2, 3; fiui = −12, −30, 0, 16, 8, 12; sum = −6. Mean = 8.5 + 3 × (−6/100) = 8.5 − 0.18 = 8.32.
Mode: modal class 7 - 10, f1 = 40, f0 = 30, f2 = 16. Mode = 7 + [10/(80 − 30 − 16)] × 3 = 7 + 30/34 = 7 + 0.88 = 7.88.
Q12. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure: 1000 - 1500 (24), 1500 - 2000 (40), 2000 - 2500 (33), 2500 - 3000 (28), 3000 - 3500 (30), 3500 - 4000 (22), 4000 - 4500 (16), 4500 - 5000 (7). (NCERT Exercise 13.2, Q3)
Answer: Mode: modal class 1500 - 2000, f1 = 40, f0 = 24, f2 = 33, h = 500. Mode = 1500 + [16/(80 − 24 − 33)] × 500 = 1500 + 8000/23 = 1500 + 347.83 = ₹ 1847.83.
Mean: class marks 1250, 1750, …, 4750. Take a = 3250, h = 500: ui = −4, −3, −2, −1, 0, 1, 2, 3. fiui = −96, −120, −66, −28, 0, 22, 32, 21; sum = −310 + 75 = −235. Mean = 3250 + 500 × (−235/200) = 3250 − 587.5 = ₹ 2662.50.
Q13. The median of the following data is 525. Find the values of x and y, if the total frequency is 100. 0 - 100 (2), 100 - 200 (5), 200 - 300 (x), 300 - 400 (12), 400 - 500 (17), 500 - 600 (20), 600 - 700 (y), 700 - 800 (9), 800 - 900 (7), 900 - 1000 (4). (NCERT Example 8)
Answer: Sum of frequencies = 76 + x + y = 100, so x + y = 24. Median 525 lies in 500 - 600, so l = 500, f = 20, h = 100, and cf = 2 + 5 + x + 12 + 17 = 36 + x. 525 = 500 + [(50 − 36 − x)/20] × 100 ⇒ 25 = (14 − x) × 5 ⇒ 25 = 70 − 5x ⇒ x = 9. Then y = 24 − 9 = 15.
Common Mistakes and Examiner Tips
- Using the class limit instead of the class mark in the mean. Fix: xi is always the mid-point, (lower + upper)/2.
- Dividing by the number of classes. Fix: divide by Σfi, the total frequency, never by how many classes there are.
- Forgetting to multiply by h in the step-deviation method. Fix: write x̄ = a + h × (Σfiui/Σfi) in full before substituting.
- Sign slips in the deviation columns. Fix: deviations below a are negative. Add the negative and positive fiui separately, then combine.
- Taking f0 and f2 from the wrong classes. Fix: f0 is the class just before the modal class and f2 the class just after. If the modal class is the first class, f0 = 0.
- Writing the mode denominator as f1 − f0 − f2. Fix: it is 2f1 − f0 − f2.
- Using the cf of the median class itself. Fix: cf is the cumulative frequency of the class preceding the median class. A median outside the median class signals this error.
- Choosing the median class by the highest frequency. Fix: the median class is chosen by cumulative frequency (first cf greater than n/2). The highest frequency locates the modal class.
- Applying the mode or median formula to classes with gaps. Fix: convert 118 - 126, 127 - 135, … to 117.5 - 126.5, 126.5 - 135.5, … first. The new l and h come from the converted classes.
- Reading a cumulative table as a frequency table. Fix: when a question says “less than” or “below”, subtract consecutive values to recover the class frequencies before doing anything else.
- Stopping at a number. Fix: when a question says “interpret” or “compare”, add one sentence with units.
Examiner tip: draw the full table with headings and a totals row, and list l, cf, f, h (or l, f0, f1, f2, h) before substituting.
Quick Revision Points
- Grouped data assumes the frequency of each class is centred at its mid-point, the class mark.
- Class mark = (upper limit + lower limit)/2.
- Direct method: x̄ = Σfixi/Σfi.
- Assumed mean method: x̄ = a + Σfidi/Σfi, with di = xi − a.
- Step-deviation method: x̄ = a + h(Σfiui/Σfi), with ui = (xi − a)/h.
- All three methods give the same mean; the mean does not depend on the choice of a.
- Mode = l + [(f1 − f0)/(2f1 − f0 − f2)] × h.
- If f0 = f2, the mode equals the class mark of the modal class.
- Cumulative frequency = frequency of the class plus the frequencies of all classes before it.
- Less than type uses upper limits and rises to n; more than type uses lower limits and falls from n.
- Median class = first class whose cumulative frequency is greater than n/2.
- Median = l + [(n/2 − cf)/f] × h, where cf belongs to the preceding class.
- Mode and median formulas need continuous classes; convert gaps using half the gap.
- Median: best for a typical value when extreme values exist, such as wages.
- Mode: best for the most popular item.
- Empirical relation: 3 Median = Mode + 2 Mean.
Weightage in Board Exams
Check the current CBSE Class 10 Maths course structure for the marks given to the unit that contains this chapter. The chapter itself is assessed in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Class mark, ui for a given xi, the empirical relation, which limits a cumulative table uses, which measure suits a situation |
| Short answers | Mean by a chosen method, mode of a distribution, converting a less than table to frequencies |
| Long answers | Median with missing frequencies, or mean, median and mode of one distribution with a comparison |
| Case-based | A survey table with parts on the modal class, the median class and one calculation |
Practise every NCERT exercise question in 13.1, 13.2 and 13.3, then the missing frequency and less than table types.
Class 10 Maths Β· Chapter 13 β swipe through all 9 cards to understand the whole chapter.
Class Mark
The mid-point that represents every observation in a class.
Assumption: the frequency of each class is centred at its mid-point.
- Class 10 – 25 has class mark 17.5.
- A value equal to an upper limit goes in the next class.
- Class size h = upper limit – lower limit.
Direct Method
Multiply each class mark by its frequency, add, and divide by total frequency.
Best when class marks and frequencies are small.
- NCERT marks data: 1860 / 30 = 62.
- Divide by Ξ£fα΅’, never by the number of classes.
- The grouped mean is approximate; the raw data gave 59.3.
Assumed Mean Method
Measure each class mark from a chosen central value a.
The mean does not depend on the choice of a.
- Pick a class mark near the middle as a.
- Marks data with a = 47.5: 47.5 + 435/30 = 62.
- Deviations below a are negative.
Step-deviation Method
Divide each deviation by the class size to get small whole numbers.
Works for unequal classes if h divides every deviation.
- Marks data: 47.5 + 15 Γ 29/30 = 62.
- Never forget to multiply by h.
- Wickets example: 200 + 20 Γ (-106/45) = 152.89.
Mode of Grouped Data
The mode lies in the modal class, the class of highest frequency.
f0 = class before, f2 = class after the modal class.
- Family sizes: 3 + (1/7) Γ 2 = 3.286.
- If f0 = f2, the mode equals the class mark.
- Denominator is 2f1 – f0 – f2.
Cumulative Frequency
Add the frequencies of all preceding classes to the class frequency.
Less than type uses upper limits; more than type uses lower limits.
- Less than type rises and ends at n.
- More than or equal to type starts at n and falls.
- Subtract consecutive values to recover class frequencies.
Median of Grouped Data
The median class is the first class whose cumulative frequency exceeds n/2.
cf is the cumulative frequency of the class before the median class.
- 53 students: 60 + (4.5/7) Γ 10 = 66.4.
- About half the data lies below the median.
- Median outside its class means a wrong cf.
Continuous Classes
Mode and median formulas need classes with no gaps between them.
Class marks do not change, so the mean is unaffected.
- 118 – 126 becomes 117.5 – 126.5.
- Take l and h from the converted classes.
- Leaves example: median = 146.75 mm.
Mean, Median or Mode
Each measure answers a different question about the data.
The relation is empirical: it links the three measures approximately.
- Mean: uses all data, good for comparing schools.
- Median: typical wage when extreme values exist.
- Mode: most popular TV programme or item.
π Practice Statistics - 10 board questions
CBSE previous-year and competency-based Β· with answers & explanations
Start βClose β
Why C: Step 1 (step-deviation formula): xΜ = a + h Ε«, so 64 = 62.5 + 5Ε«. Step 2 (solve): 5Ε« = 1.5, Ε« = 0.3.
Why not A: 0.5 fails the check, since 62.5 + 5 Γ 0.5 = 65, not 64.
Why not B: 1.5 is xΜ β a = dΜ, the mean deviation before dividing by h.
Why not D: 7.5 multiplies 1.5 by 5 instead of dividing.
Remember: Ε« = (xΜ β a)/h, the same scaling as each u = (x β a)/h.
Why C: Step 1 (empirical relation): 3 Median = Mode + 2 Mean = 21 + 2 Γ 12 = 45. Step 2 (divide by 3): Median = 45/3 = 15.
Why not A: 6 fails the check, since 3 Γ 6 = 18, not 45; it is what swapping the coefficients gives (3 Γ 12 β 2 Γ 15 = 6).
Why not B: 13.5 fails the check, since 3 Γ 13.5 = 40.5, not 45; it is the plain average of 12 and 15, which ignores the weights in the relation.
Why not D: 14 = (2 Γ 21)/3 puts the factor 2 on the mode and drops the mean; 3 Γ 14 = 42, not 45.
Remember: Median = (Mode + 2 Mean)/3, and it lies between the mean and the mode.
Why A: Step 1 (empirical relation): Mode = 3 Median β 2 Mean. Step 2 (substitute): Mode = 3 Γ 40 β 2 Γ 43 = 120 β 86 = 34.
Why not B: 43 is the mean itself; the mode need not equal the mean.
Why not C: 38.5 = 40 β (43 β 40)/2 uses half the gap, but the relation gives Mode β Median = 2(Median β Mean).
Why not D: 41.5 is the plain average of 43 and 40, which ignores the relation.
Remember: the mode lies on the far side of the median from the mean, twice as far.
Why D: Step 1 (sum = mean Γ n): there are 7 observations, so the sum = 7 Γ 7 = 49. Step 2 (add the terms): 2 + 9 + (x + 6) + (2x + 3) + 5 + 10 + 5 = 40 + 3x. So 40 + 3x = 49, 3x = 9, x = 3.
Why not A: 9 is 3x, the step before dividing by 3.
Why not B: 6 drops the constant terms 6 and 3 from x + 6 and 2x + 3, giving 31 + 3x = 49.
Why not C: 5 fits neither total; with x = 5 the sum is 55, so the mean would be 55/7, not 7.
Remember: mean Γ number of observations = sum; set up that equation first.
Why C: Step 1 (empirical relation): 3 Median = Mode + 2 Mean = 15x + 36x = 51x. Step 2 (divide by 3): Median = 17x.
Why not A: x comes from subtracting instead of adding, (18x β 15x)/3.
Why not B: 11x = (15x + 18x)/3 drops the factor 2 on the mean.
Why not D: 34x is double the correct value; 3 Γ 34x = 102x, not 51x.
Remember: Median = (Mode + 2 Mean)/3, so it lies between the mode and the mean.
Why A: Adding 2 to every value keeps the order of the observations unchanged, so the same observation stays in the middle, and that middle value is now 2 more. So the median increases by 2.
Why not B: 2n would be the rise in the total of all values; the median is one middle value, not a total.
Why not C: the middle position is the same, but the value sitting there has gone up.
Why not D: the values were increased, so the median cannot go down.
Remember: shift every value by k and the mean, median and mode all shift by k.
Why C: Step 1 (sum = mean Γ n): all five add to 5 Γ 15 = 75; the first three add to 3 Γ 14 = 42; the last three add to 3 Γ 17 = 51. Step 2 (overlap counted twice): 42 + 51 = 93 counts the third observation twice, so the third observation = 93 β 75 = 18.
Why not A: with 20 as the shared value the two groups would total 75 + 20 = 95, but they total 93.
Why not B: with 19 the two groups would total 75 + 19 = 94, not 93.
Why not D: 17 is just the mean of the last three, read off as if it were the third value.
Remember: when two groups share an observation, (sum of both groups) β (total) = the shared value.
Why A: Step 1 (empirical relation): 3 Median = Mode + 2 Mean, so Mode β Median = 2 Median β 2 Mean = 2(Median β Mean). Step 2 (substitute): 24 = 2(Median β Mean), so Median β Mean = 12.
Why not B: this assumes the two differences are equal, but the relation makes one twice the other.
Why not C: 8 divides 24 by 3, using the coefficient of the median instead of the factor 2.
Why not D: 36 = 24 + 12 is Mode β Mean, the whole gap, not Median β Mean.
Remember: Mode β Median = 2(Median β Mean); the median sits one third of the way from the mean to the mode.
Why B: The marks obtained by the maximum number of students is the value that occurs most often, and NCERT defines the mode as the value that occurs most frequently.
Why not A: the median is the middle value after arranging marks in order, not the most common one.
Why not C: the mean is the average of all marks and need not be a mark that anyone scored most often.
Why not D: range measures spread (highest minus lowest), not a typical or most common value.
Remember: most students, most common, most popular: mode.
Why A: Step 1 (sum of observations): each of the n terms drops by 2, so the sum drops by 2n. Step 2 (mean = sum/n): the new mean = (old sum β 2n)/n = old mean β 2.
Why not B: every value changed, so the average must change too.
Why not C: 2n is the fall in the SUM; dividing by n brings it back to 2.
Why not D: there is no halving anywhere; the whole shift of 2 passes to the mean.
Remember: add or subtract a constant to every value and the mean moves by exactly that constant.
Chapter Navigation
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- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
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- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
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Frequently Asked Questions
The class mark is the mid-point of a class, (upper limit + lower limit)/2. In grouped data the individual observations are not known, so NCERT assumes the frequency of each class is centred around its mid-point. The class mark then stands for every observation in that class and is used as x in the mean formulas.
All three methods give the same mean. Use the direct method when the class marks and frequencies are small. Use the assumed mean method when the class marks are large. Use the step-deviation method when the deviations from the assumed mean share a common factor, usually the class size. It also works for unequal classes if h divides every deviation.
Grouping replaces every observation in a class by the class mark. In NCERT’s example, 30 students’ marks give an exact mean of 59.3 when taken one by one, but 62 after grouping into classes of width 15. The difference comes from the mid-point assumption, so the grouped mean is an approximate mean.
The modal class is the class with the highest frequency. Note its lower limit l, the class size h, its frequency f1, and the frequencies f0 and f2 of the classes just before and after it. Then Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] Γ h. If f0 equals f2, the mode is exactly the class mark.
Find n, the total frequency, and n/2. Write the cumulative frequency column. The median class is the first class whose cumulative frequency is greater than n/2. In the formula Median = l + [(n/2 - cf)/f] Γ h, cf is the cumulative frequency of the class before the median class, never of the median class itself.
A less than type table counts observations below each upper class limit, so it rises and ends at n. A more than or equal to type table counts observations at or above each lower class limit, so it starts at n and falls. To get class frequencies back from either table, subtract consecutive cumulative frequencies.
These classes have gaps, and the mode and median formulas need continuous classes. Take half the gap, here 0.5, subtract it from every lower limit and add it to every upper limit. The classes become 117.5 - 126.5, 126.5 - 135.5, and so on. Class marks stay the same, so the mean is unaffected.
3 Median = Mode + 2 Mean. It lets you find one measure when the other two are known. For example, if the median is 26 and the mean is 27, the mode is 3 Γ 26 - 2 Γ 27 = 24. Rearranged forms are Mode = 3 Median - 2 Mean and Mean = (3 Median - Mode)/2.
