Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 13: Statistics
Chapter 13 covers three measures of central tendency for grouped data - mean (direct method, assumed mean method, step deviation method), mode, and median. Students also learn to represent cumulative frequency distributions graphically as ogives (less than type and more than type) and use them to find the median. This chapter is part of the Statistics and Probability unit which carries 11 marks in the CBSE Class 10 board exam 2027. It is one of the most scoring chapters as all questions follow a fixed formula-based pattern.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 13 Statistics has 4 exercises: Exercise 13.1 (9 questions on mean of grouped data), Exercise 13.2 (6 questions on mode of grouped data), Exercise 13.3 (7 questions on median of grouped data), and Exercise 13.4 (3 questions on cumulative frequency graphs/ogives). All 4 exercises are part of the CBSE board exam syllabus. The empirical relationship 3 Median = Mode + 2 Mean is not in the NCERT textbook but is frequently asked in board exams and competitive exams.
Table of Contents
Exercise 13.1 - Mean of Grouped Data
9 questions
Exercise 13.2 - Mode of Grouped Data
6 questions
Exercise 13.3 - Median of Grouped Data
7 questions
Exercise 13.4 - Graphical Representation of Cumulative Frequency Distribution (Ogive)
3 questions
Previous Year Questions
15 questions (CBSE 2019-2026)
MCQ Practice
12 questions
13.1Exercise 13.1 - Mean of Grouped Data
Q1A 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
Using the direct method:
| Class | fi | xi (mid-point) | fi x xi |
|---|---|---|---|
| 0-2 | 1 | 1 | 1 |
| 2-4 | 2 | 3 | 6 |
| 4-6 | 1 | 5 | 5 |
| 6-8 | 5 | 7 | 35 |
| 8-10 | 6 | 9 | 54 |
| 10-12 | 2 | 11 | 22 |
| 12-14 | 3 | 13 | 39 |
Sum of fi = 20, Sum of fi x xi = 162
Mean = (Sum of fi x xi) / (Sum of fi) = 162/20 = 8.1 plants per house
Q2Consider the following distribution of daily wages of 50 workers of a factory. Daily wages (in Rs): 500-520, 520-540, 540-560, 560-580, 580-600 Number of workers: 12, 14, 8, 6, 10 Find the mean daily wages of the workers of the factory by using an appropriate method.
Using step deviation method with assumed mean a = 550 and h = 20:
| Class | fi | xi | ui = (xi - 550)/20 | fi x ui |
|---|---|---|---|---|
| 500-520 | 12 | 510 | -2 | -24 |
| 520-540 | 14 | 530 | -1 | -14 |
| 540-560 | 8 | 550 | 0 | 0 |
| 560-580 | 6 | 570 | 1 | 6 |
| 580-600 | 10 | 590 | 2 | 20 |
Sum of fi = 50, Sum of fi x ui = -12
Mean = a + h x (Sum of fi x ui)/(Sum of fi)
= 550 + 20 x (-12/50)
= 550 + 20 x (-0.24)
= 550 - 4.8
= Rs 545.20
Q3The following table gives the distribution of the life time of 400 neon lamps: Life time (in hours): 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Number of lamps: 14, 56, 60, 86, 74, 62, 48
Using step deviation method with assumed mean a = 3250 and h = 500:
| Class | fi | xi | ui = (xi - 3250)/500 | fi x ui |
|---|---|---|---|---|
| 1500-2000 | 14 | 1750 | -3 | -42 |
| 2000-2500 | 56 | 2250 | -2 | -112 |
| 2500-3000 | 60 | 2750 | -1 | -60 |
| 3000-3500 | 86 | 3250 | 0 | 0 |
| 3500-4000 | 74 | 3750 | 1 | 74 |
| 4000-4500 | 62 | 4250 | 2 | 124 |
| 4500-5000 | 48 | 4750 | 3 | 144 |
Sum of fi = 400, Sum of fi x ui = 128
Mean = a + h x (Sum of fi x ui)/(Sum of fi)
= 3250 + 500 x (128/400)
= 3250 + 500 x 0.32
= 3250 + 160
= 3410 hours
Q4The following table gives the frequency distribution of the number of orders received at each day during a month at a certain company: Number of orders: 10-12, 12-14, 14-16, 16-18, 18-20 Number of days: 4, 12, 5, 2, 2 (i) Find the mean daily orders. (ii) Find the mean daily orders by using the assumed mean method. (iii) Find the mean daily orders by using the step deviation method.
(i) Direct method:
| Class | fi | xi | fi x xi |
|---|---|---|---|
| 10-12 | 4 | 11 | 44 |
| 12-14 | 12 | 13 | 156 |
| 14-16 | 5 | 15 | 75 |
| 16-18 | 2 | 17 | 34 |
| 18-20 | 2 | 19 | 38 |
Sum of fi = 25, Sum of fi x xi = 347
Mean = 347/25 = 13.88 orders (approx.)
(ii) Assumed mean method (a = 13):
| Class | fi | xi | di = xi - 13 | fi x di |
|---|---|---|---|---|
| 10-12 | 4 | 11 | -2 | -8 |
| 12-14 | 12 | 13 | 0 | 0 |
| 14-16 | 5 | 15 | 2 | 10 |
| 16-18 | 2 | 17 | 4 | 8 |
| 18-20 | 2 | 19 | 6 | 12 |
Sum of fi = 25, Sum of fi x di = 22
Mean = 13 + 22/25 = 13 + 0.88 = 13.88 orders
(iii) Step deviation method (a = 13, h = 2):
| Class | fi | xi | ui = (xi - 13)/2 | fi x ui |
|---|---|---|---|---|
| 10-12 | 4 | 11 | -1 | -4 |
| 12-14 | 12 | 13 | 0 | 0 |
| 14-16 | 5 | 15 | 1 | 5 |
| 16-18 | 2 | 17 | 2 | 4 |
| 18-20 | 2 | 19 | 3 | 6 |
Sum of fi = 25, Sum of fi x ui = 11
Mean = 13 + 2 x (11/25) = 13 + 2 x 0.44 = 13 + 0.88 = 13.88 orders
All three methods give the same answer: 13.88 orders per day.
Q5The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Daily pocket allowance (in Rs): 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25 Number of children: 7, 6, 9, 13, f, 5, 4
Using direct method:
| Class | fi | xi | fi x xi |
|---|---|---|---|
| 11-13 | 7 | 12 | 84 |
| 13-15 | 6 | 14 | 84 |
| 15-17 | 9 | 16 | 144 |
| 17-19 | 13 | 18 | 234 |
| 19-21 | f | 20 | 20f |
| 21-23 | 5 | 22 | 110 |
| 23-25 | 4 | 24 | 96 |
Sum of fi = 44 + f
Sum of fi x xi = 752 + 20f
Mean = (Sum of fi x xi)/(Sum of fi)
18 = (752 + 20f)/(44 + f)
18(44 + f) = 752 + 20f
792 + 18f = 752 + 20f
792 - 752 = 20f - 18f
40 = 2f
f = 20
The missing frequency f = 20
Q630 women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. Number of heartbeats per minute: 65-68, 68-71, 71-74, 74-77, 77-80, 80-83, 83-86 Number of women: 2, 4, 3, 8, 7, 4, 2
Using step deviation method with assumed mean a = 75.5 and h = 3:
| Class | fi | xi | ui = (xi - 75.5)/3 | fi x ui |
|---|---|---|---|---|
| 65-68 | 2 | 66.5 | -3 | -6 |
| 68-71 | 4 | 69.5 | -2 | -8 |
| 71-74 | 3 | 72.5 | -1 | -3 |
| 74-77 | 8 | 75.5 | 0 | 0 |
| 77-80 | 7 | 78.5 | 1 | 7 |
| 80-83 | 4 | 81.5 | 2 | 8 |
| 83-86 | 2 | 84.5 | 3 | 6 |
Sum of fi = 30, Sum of fi x ui = 4
Mean = a + h x (Sum of fi x ui)/(Sum of fi)
= 75.5 + 3 x (4/30)
= 75.5 + 3 x 0.1333
= 75.5 + 0.4
= 75.9 beats per minute
Q7In 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
First, the class intervals are not continuous. Converting to continuous form by adjusting boundaries by 0.5:
| Original Class | Adjusted Class | fi | xi |
|---|---|---|---|
| 50-52 | 49.5-52.5 | 15 | 51 |
| 53-55 | 52.5-55.5 | 110 | 54 |
| 56-58 | 55.5-58.5 | 135 | 57 |
| 59-61 | 58.5-61.5 | 115 | 60 |
| 62-64 | 61.5-64.5 | 25 | 63 |
Using step deviation method with a = 57, h = 3:
| Class | fi | xi | ui = (xi - 57)/3 | fi x ui |
|---|---|---|---|---|
| 49.5-52.5 | 15 | 51 | -2 | -30 |
| 52.5-55.5 | 110 | 54 | -1 | -110 |
| 55.5-58.5 | 135 | 57 | 0 | 0 |
| 58.5-61.5 | 115 | 60 | 1 | 115 |
| 61.5-64.5 | 25 | 63 | 2 | 50 |
Sum of fi = 400, Sum of fi x ui = 25
Mean = a + h x (Sum of fi x ui)/(Sum of fi)
= 57 + 3 x (25/400)
= 57 + 3 x 0.0625
= 57 + 0.1875
= 57.19 mangoes (approx.)
Q8The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure (in Rs): 100-150, 150-200, 200-250, 250-300, 300-350 Number of households: 4, 5, 12, 2, 2 Find the mean daily expenditure on food by a suitable method.
Using step deviation method with assumed mean a = 225, h = 50:
| Class | fi | xi | ui = (xi - 225)/50 | fi x ui |
|---|---|---|---|---|
| 100-150 | 4 | 125 | -2 | -8 |
| 150-200 | 5 | 175 | -1 | -5 |
| 200-250 | 12 | 225 | 0 | 0 |
| 250-300 | 2 | 275 | 1 | 2 |
| 300-350 | 2 | 325 | 2 | 4 |
Sum of fi = 25, Sum of fi x ui = -7
Mean = a + h x (Sum of fi x ui)/(Sum of fi)
= 225 + 50 x (-7/25)
= 225 + 50 x (-0.28)
= 225 - 14
= Rs 211
Q9The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55 Number of states/U.T.: 3, 8, 9, 10, 3, 0, 0, 2
Mean (using step deviation method with a = 32.5, h = 5):
| Class | fi | xi | ui = (xi - 32.5)/5 | fi x ui |
|---|---|---|---|---|
| 15-20 | 3 | 17.5 | -3 | -9 |
| 20-25 | 8 | 22.5 | -2 | -16 |
| 25-30 | 9 | 27.5 | -1 | -9 |
| 30-35 | 10 | 32.5 | 0 | 0 |
| 35-40 | 3 | 37.5 | 1 | 3 |
| 40-45 | 0 | 42.5 | 2 | 0 |
| 45-50 | 0 | 47.5 | 3 | 0 |
| 50-55 | 2 | 52.5 | 4 | 8 |
Sum of fi = 35, Sum of fi x ui = -23
Mean = 32.5 + 5 x (-23/35) = 32.5 + 5 x (-0.657) = 32.5 - 3.29 = 29.21 (approx.)
Mode:
Modal class = 30-35 (highest frequency = 10)
l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 30 + [(10 - 9)/(20 - 9 - 3)] x 5
= 30 + [1/8] x 5
= 30 + 0.625
= 30.625 (approx.)
Interpretation: The mean student-teacher ratio is about 29.2, meaning on average there are about 29 students per teacher across states. The mode is about 30.6, indicating that the most common ratio is about 30-31 students per teacher. Both measures are close, suggesting the distribution is fairly symmetric around 30 students per teacher.
13.2Exercise 13.2 - Mode of Grouped Data
Q1The following table shows the ages of the patients admitted in a hospital during a year: Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65 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 (highest frequency = 23)
l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 35 + [(23 - 21)/(46 - 21 - 14)] x 10
= 35 + [2/11] x 10
= 35 + 20/11
= 35 + 1.818
= 36.82 years (approx.)
Mean (using assumed mean method, a = 30, h = 10):
| Class | fi | xi | di = xi - 30 | fi x di |
|---|---|---|---|---|
| 5-15 | 6 | 10 | -20 | -120 |
| 15-25 | 11 | 20 | -10 | -110 |
| 25-35 | 21 | 30 | 0 | 0 |
| 35-45 | 23 | 40 | 10 | 230 |
| 45-55 | 14 | 50 | 20 | 280 |
| 55-65 | 5 | 60 | 30 | 150 |
Sum of fi = 80, Sum of fi x di = 430
Mean = a + (Sum of fi x di)/(Sum of fi) = 30 + 430/80 = 30 + 5.375 = 35.38 years (approx.)
Interpretation: The mode (36.82 years) tells us that the maximum number of patients admitted in the hospital are in the age group 35-45. The mean (35.38 years) gives us the average age. Since both values are close, the data is fairly symmetric, centred around 35-37 years.
Q2The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Lifetimes (in hours): 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 Frequency: 10, 35, 52, 61, 38, 29 Determine the modal lifetimes of the components.
Modal class = 60-80 (highest frequency = 61)
l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 60 + [(61 - 52)/(122 - 52 - 38)] x 20
= 60 + [9/32] x 20
= 60 + 180/32
= 60 + 5.625
= 65.625 hours
The modal lifetime of the electrical components is approximately 65.63 hours.
Q3The 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. Expenditure (in Rs): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Number of families: 24, 40, 33, 28, 30, 22, 16, 7
Mode:
Modal class = 1500-2000 (highest frequency = 40)
l = 1500, f1 = 40, f0 = 24, f2 = 33, h = 500
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 1500 + [(40 - 24)/(80 - 24 - 33)] x 500
= 1500 + [16/23] x 500
= 1500 + 8000/23
= 1500 + 347.83
= Rs 1847.83 (approx.)
Mean (using step deviation method, a = 2750, h = 500):
| Class | fi | xi | ui = (xi - 2750)/500 | fi x ui |
|---|---|---|---|---|
| 1000-1500 | 24 | 1250 | -3 | -72 |
| 1500-2000 | 40 | 1750 | -2 | -80 |
| 2000-2500 | 33 | 2250 | -1 | -33 |
| 2500-3000 | 28 | 2750 | 0 | 0 |
| 3000-3500 | 30 | 3250 | 1 | 30 |
| 3500-4000 | 22 | 3750 | 2 | 44 |
| 4000-4500 | 16 | 4250 | 3 | 48 |
| 4500-5000 | 7 | 4750 | 4 | 28 |
Sum of fi = 200, Sum of fi x ui = -35
Mean = 2750 + 500 x (-35/200) = 2750 + 500 x (-0.175) = 2750 - 87.5 = Rs 2662.50
Q4The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55 Number of states/U.T.: 3, 8, 9, 10, 3, 0, 0, 2
Mode:
Modal class = 30-35 (highest frequency = 10)
l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 30 + [(10 - 9)/(20 - 9 - 3)] x 5
= 30 + [1/8] x 5
= 30 + 0.625
= 30.6 (approx.)
Mean (using step deviation method, a = 32.5, h = 5):
| Class | fi | xi | ui = (xi - 32.5)/5 | fi x ui |
|---|---|---|---|---|
| 15-20 | 3 | 17.5 | -3 | -9 |
| 20-25 | 8 | 22.5 | -2 | -16 |
| 25-30 | 9 | 27.5 | -1 | -9 |
| 30-35 | 10 | 32.5 | 0 | 0 |
| 35-40 | 3 | 37.5 | 1 | 3 |
| 40-45 | 0 | 42.5 | 2 | 0 |
| 45-50 | 0 | 47.5 | 3 | 0 |
| 50-55 | 2 | 52.5 | 4 | 8 |
Sum of fi = 35, Sum of fi x ui = -23
Mean = 32.5 + 5 x (-23/35) = 32.5 - 3.29 = 29.2 (approx.)
Interpretation: The mode (30.6) indicates that the most common student-teacher ratio is about 30-31 students per teacher. The mean (29.2) shows the overall average is about 29 students per teacher. Both values are close, indicating a fairly uniform distribution centred around 30 students per teacher across Indian states.
Q5A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table below. Find the mode of the data. Number of cars: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80 Frequency: 7, 14, 13, 12, 20, 11, 15, 8
Modal class = 40-50 (highest frequency = 20)
l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 40 + [(20 - 12)/(40 - 12 - 11)] x 10
= 40 + [8/17] x 10
= 40 + 80/17
= 40 + 4.706
= 44.7 cars (approx.)
Q6A student noted the number of phone calls received by a telephone operator for 100 days in a locality and prepared the following frequency distribution. Find the mode. Number of calls: 10-12, 12-14, 14-16, 16-18, 18-20, 20-22 Number of days: 14, 21, 28, 16, 10, 11
Modal class = 14-16 (highest frequency = 28)
l = 14, f1 = 28, f0 = 21, f2 = 16, h = 2
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
= 14 + [(28 - 21)/(56 - 21 - 16)] x 2
= 14 + [7/19] x 2
= 14 + 14/19
= 14 + 0.7368
= 14.74 calls (approx.)
13.3Exercise 13.3 - Median of Grouped Data
Q1The 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. Monthly consumption (in units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205 Number of consumers: 4, 5, 13, 20, 14, 8, 4
Median:
n = 68, n/2 = 34
Cumulative frequency table:
| Class | fi | cf |
|---|---|---|
| 65-85 | 4 | 4 |
| 85-105 | 5 | 9 |
| 105-125 | 13 | 22 |
| 125-145 | 20 | 42 |
| 145-165 | 14 | 56 |
| 165-185 | 8 | 64 |
| 185-205 | 4 | 68 |
Median class = 125-145 (cf just exceeds n/2 = 34)
l = 125, cf = 22, f = 20, h = 20
Median = l + [(n/2 - cf)/f] x h
= 125 + [(34 - 22)/20] x 20
= 125 + [12/20] x 20
= 125 + 12
= 137 units
Mean (using step deviation, a = 135, h = 20):
| Class | fi | xi | ui = (xi - 135)/20 | fi x ui |
|---|---|---|---|---|
| 65-85 | 4 | 75 | -3 | -12 |
| 85-105 | 5 | 95 | -2 | -10 |
| 105-125 | 13 | 115 | -1 | -13 |
| 125-145 | 20 | 135 | 0 | 0 |
| 145-165 | 14 | 155 | 1 | 14 |
| 165-185 | 8 | 175 | 2 | 16 |
| 185-205 | 4 | 195 | 3 | 12 |
Sum of fi = 68, Sum of fi x ui = 7
Mean = 135 + 20 x (7/68) = 135 + 2.06 = 137.06 units (approx.)
Mode:
Modal class = 125-145 (highest frequency = 20)
l = 125, f1 = 20, f0 = 13, f2 = 14, h = 20
Mode = 125 + [(20 - 13)/(40 - 13 - 14)] x 20 = 125 + [7/13] x 20 = 125 + 10.77 = 135.77 units (approx.)
Comparison: Mean (137.06), Median (137), Mode (135.77) are all very close, indicating the data is approximately symmetric.
Q2If the median of the distribution given below is 28.5, find the values of x and y. Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Frequency: 5, x, 20, 15, y, 5 Total frequency = 60
Total frequency: 5 + x + 20 + 15 + y + 5 = 60
So x + y = 15 ... (i)
n = 60, n/2 = 30
Median = 28.5, so the median class is 20-30.
Cumulative frequencies:
| Class | fi | cf |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | x | 5 + x |
| 20-30 | 20 | 25 + x |
| 30-40 | 15 | 40 + x |
| 40-50 | y | 40+x+y |
| 50-60 | 5 | 60 |
For the median class 20-30:
l = 20, cf = 5 + x, f = 20, h = 10
Median = l + [(n/2 - cf)/f] x h
28.5 = 20 + [(30 - (5 + x))/20] x 10
28.5 = 20 + [(25 - x)/20] x 10
28.5 = 20 + (25 - x)/2
8.5 = (25 - x)/2
17 = 25 - x
x = 8
From (i): y = 15 - 8 = 7
x = 8, y = 7
Q3A 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 years. Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60 Number of policy holders: 2, 6, 24, 45, 78, 89, 92, 98, 100
First, convert to class intervals and find frequencies:
| Class | cf | fi |
|---|---|---|
| 15-20 | 2 | 2 |
| 20-25 | 6 | 6-2 = 4 |
| 25-30 | 24 | 24-6 = 18 |
| 30-35 | 45 | 45-24 = 21 |
| 35-40 | 78 | 78-45 = 33 |
| 40-45 | 89 | 89-78 = 11 |
| 45-50 | 92 | 92-89 = 3 |
| 50-55 | 98 | 98-92 = 6 |
| 55-60 | 100 | 100-98 = 2 |
n = 100, n/2 = 50
Median class = 35-40 (cf = 78 is the first to exceed 50)
l = 35, cf = 45 (cf of class before median class), f = 33, h = 5
Median = l + [(n/2 - cf)/f] x h
= 35 + [(50 - 45)/33] x 5
= 35 + [5/33] x 5
= 35 + 25/33
= 35 + 0.76
= 35.76 years (approx.)
Q4The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: Length (in mm): 118-126, 127-135, 136-144, 145-153, 154-162, 163-171, 172-180 Number of leaves: 3, 5, 9, 12, 5, 4, 2 Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes.)
The classes are not continuous. Converting to continuous form by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
| Adjusted Class | fi | cf |
|---|---|---|
| 117.5-126.5 | 3 | 3 |
| 126.5-135.5 | 5 | 8 |
| 135.5-144.5 | 9 | 17 |
| 144.5-153.5 | 12 | 29 |
| 153.5-162.5 | 5 | 34 |
| 162.5-171.5 | 4 | 38 |
| 171.5-180.5 | 2 | 40 |
n = 40, n/2 = 20
Median class = 144.5-153.5 (cf = 29 is the first to exceed 20)
l = 144.5, cf = 17, f = 12, h = 9
Median = l + [(n/2 - cf)/f] x h
= 144.5 + [(20 - 17)/12] x 9
= 144.5 + [3/12] x 9
= 144.5 + 0.25 x 9
= 144.5 + 2.25
= 146.75 mm
Q5The following table gives the distribution of the life time of 400 neon lamps: Life time (in hours): 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Number of lamps: 14, 56, 60, 86, 74, 62, 48 Find the median life time of a lamp.
Cumulative frequency table:
| Class | fi | cf |
|---|---|---|
| 1500-2000 | 14 | 14 |
| 2000-2500 | 56 | 70 |
| 2500-3000 | 60 | 130 |
| 3000-3500 | 86 | 216 |
| 3500-4000 | 74 | 290 |
| 4000-4500 | 62 | 352 |
| 4500-5000 | 48 | 400 |
n = 400, n/2 = 200
Median class = 3000-3500 (cf = 216 is the first to exceed 200)
l = 3000, cf = 130, f = 86, h = 500
Median = l + [(n/2 - cf)/f] x h
= 3000 + [(200 - 130)/86] x 500
= 3000 + [70/86] x 500
= 3000 + 0.8139 x 500
= 3000 + 406.98
= 3406.98 hours (approx.)
Q6100 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, 4-7, 7-10, 10-13, 13-16, 16-19 Number of surnames: 6, 30, 40, 16, 4, 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.
Median:
| Class | fi | cf |
|---|---|---|
| 1-4 | 6 | 6 |
| 4-7 | 30 | 36 |
| 7-10 | 40 | 76 |
| 10-13 | 16 | 92 |
| 13-16 | 4 | 96 |
| 16-19 | 4 | 100 |
n = 100, n/2 = 50
Median class = 7-10 (cf = 76 is the first to exceed 50)
l = 7, cf = 36, f = 40, h = 3
Median = 7 + [(50 - 36)/40] x 3 = 7 + [14/40] x 3 = 7 + 1.05 = 8.05 letters
Mean (using step deviation, a = 8.5, h = 3):
| Class | fi | xi | ui = (xi - 8.5)/3 | fi x ui |
|---|---|---|---|---|
| 1-4 | 6 | 2.5 | -2 | -12 |
| 4-7 | 30 | 5.5 | -1 | -30 |
| 7-10 | 40 | 8.5 | 0 | 0 |
| 10-13 | 16 | 11.5 | 1 | 16 |
| 13-16 | 4 | 14.5 | 2 | 8 |
| 16-19 | 4 | 17.5 | 3 | 12 |
Sum of fi = 100, Sum of fi x ui = -6
Mean = 8.5 + 3 x (-6/100) = 8.5 - 0.18 = 8.32 letters
Mode:
Modal class = 7-10 (highest frequency = 40)
l = 7, f1 = 40, f0 = 30, f2 = 16, h = 3
Mode = 7 + [(40 - 30)/(80 - 30 - 16)] x 3 = 7 + [10/34] x 3 = 7 + 0.88 = 7.88 letters
Q7The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg): 40-45, 45-50, 50-55, 55-60, 60-65, 65-70, 70-75 Number of students: 2, 3, 8, 6, 6, 3, 2
Cumulative frequency table:
| Class | fi | cf |
|---|---|---|
| 40-45 | 2 | 2 |
| 45-50 | 3 | 5 |
| 50-55 | 8 | 13 |
| 55-60 | 6 | 19 |
| 60-65 | 6 | 25 |
| 65-70 | 3 | 28 |
| 70-75 | 2 | 30 |
n = 30, n/2 = 15
Median class = 55-60 (cf = 19 is the first to exceed 15)
l = 55, cf = 13, f = 6, h = 5
Median = l + [(n/2 - cf)/f] x h
= 55 + [(15 - 13)/6] x 5
= 55 + [2/6] x 5
= 55 + 10/6
= 55 + 1.67
= 56.67 kg (approx.)
13.4Exercise 13.4 - Graphical Representation of Cumulative Frequency Distribution (Ogive)
Q1The following distribution gives the daily income of 50 workers of a factory. Daily income (in Rs): 100-120, 120-140, 140-160, 160-180, 180-200 Number of workers: 12, 14, 8, 6, 10 Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Less than type cumulative frequency table:
| Upper class boundary | Cumulative Frequency |
|---|---|
| Less than 120 | 12 |
| Less than 140 | 12 + 14 = 26 |
| Less than 160 | 26 + 8 = 34 |
| Less than 180 | 34 + 6 = 40 |
| Less than 200 | 40 + 10 = 50 |
Steps to draw the ogive:
- Mark the upper class boundaries (120, 140, 160, 180, 200) on the x-axis.
- Mark the cumulative frequencies on the y-axis.
- Plot the points: (120, 12), (140, 26), (160, 34), (180, 40), (200, 50).
- Also plot the starting point (100, 0).
- Join these points by a free-hand smooth curve to get the less than ogive.
Finding median from the ogive:
n/2 = 50/2 = 25
Locate 25 on the y-axis. Draw a horizontal line to meet the ogive. From the point of intersection, draw a vertical line to the x-axis. The value on the x-axis gives the median.
From the graph, median is approximately Rs 138.57.
Verification using formula: Median class = 120-140
l = 120, cf = 12, f = 14, h = 20
Median = 120 + [(25 - 12)/14] x 20 = 120 + [13/14] x 20 = 120 + 18.57 = Rs 138.57
Q2During the medical check-up of 35 students of a class, their weights were recorded as follows: Weight (in kg): Less than 38, Less than 40, Less than 42, Less than 44, Less than 46, Less than 48, Less than 50, Less than 52 Number of students: 0, 3, 5, 9, 14, 28, 32, 35 Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Steps to draw the ogive:
- Plot the points from the given less than cumulative frequency data:
(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35) - Join these points by a smooth freehand curve.
Finding median from graph:
n = 35, n/2 = 17.5
Locate 17.5 on the y-axis, draw a horizontal line to meet the ogive, then draw a vertical line to the x-axis.
From the graph, median is approximately 46.5 kg.
Verification using formula:
First, convert to frequency distribution:
| Class | fi | cf |
|---|---|---|
| 36-38 | 0 | 0 |
| 38-40 | 3-0 = 3 | 3 |
| 40-42 | 5-3 = 2 | 5 |
| 42-44 | 9-5 = 4 | 9 |
| 44-46 | 14-9 = 5 | 14 |
| 46-48 | 28-14 = 14 | 28 |
| 48-50 | 32-28 = 4 | 32 |
| 50-52 | 35-32 = 3 | 35 |
n/2 = 17.5
Median class = 46-48 (cf = 28 is the first to exceed 17.5)
l = 46, cf = 14, f = 14, h = 2
Median = 46 + [(17.5 - 14)/14] x 2
= 46 + [3.5/14] x 2
= 46 + 0.5
= 46.5 kg
The graphical and calculated values match.
Q3The following table gives production yield per hectare of wheat of 100 farms of a village. Production yield (in kg/ha): 50-55, 55-60, 60-65, 65-70, 70-75, 75-80 Number of farms: 2, 8, 12, 24, 38, 16 Change the distribution to a more than type distribution and draw its ogive.
More than type cumulative frequency table:
| Lower class boundary | Cumulative Frequency |
|---|---|
| More than or equal to 50 | 100 |
| More than or equal to 55 | 100 - 2 = 98 |
| More than or equal to 60 | 98 - 8 = 90 |
| More than or equal to 65 | 90 - 12 = 78 |
| More than or equal to 70 | 78 - 24 = 54 |
| More than or equal to 75 | 54 - 38 = 16 |
Steps to draw the ogive:
- Mark the lower class boundaries (50, 55, 60, 65, 70, 75) on the x-axis.
- Mark the cumulative frequencies on the y-axis.
- Plot the points: (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16).
- Also plot the ending point (80, 0).
- Join these points by a smooth freehand curve to get the more than ogive.
The more than ogive is a decreasing curve from left to right.
Finding median from the ogive:
n/2 = 100/2 = 50
Locate 50 on the y-axis. Draw a horizontal line to the ogive, then drop a vertical to the x-axis.
Median from graph is approximately 70.83 kg/ha.
Verification: Median class = 70-75
l = 70, cf = 62 (from less than type: 2+8+12+24 = 46... let us recalculate)
Less than cumulative frequency:
50-55: 2, 55-60: 10, 60-65: 22, 65-70: 46, 70-75: 84, 75-80: 100
n/2 = 50, Median class = 70-75
l = 70, cf = 46, f = 38, h = 5
Median = 70 + [(50 - 46)/38] x 5 = 70 + [4/38] x 5 = 70 + 0.526 = 70.53 kg/ha (approx.)
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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. Pocket allowance (Rs): 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25 Frequency: 7, 6, 9, 13, f, 5, 4
Using the direct method, we calculate the sum of fi x xi for each class. The mid-points are 12, 14, 16, 18, 20, 22, 24. Sum of fi = 44 + f and Sum of fi x xi = 752 + 20f. Setting up the equation: 18 = (752 + 20f)/(44 + f). Cross-multiplying gives 792 + 18f = 752 + 20f, which simplifies to 2f = 40, giving f = 20.
Find the mean of the following frequency distribution using step deviation method: Class: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 7, 10, 15, 8, 10
Taking assumed mean a = 25 and h = 10, mid-points are 5, 15, 25, 35, 45. The ui values are -2, -1, 0, 1, 2. Calculating fi x ui: -14, -10, 0, 8, 20. Sum of fi = 50 and Sum of fi x ui = 4. Mean = 25 + 10 x (4/50) = 25 + 0.8 = 25.8.
The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Class: 0-100, 100-200, 200-300, 300-400, 400-500, 500-600, 600-700, 700-800, 800-900, 900-1000 Frequency: 2, 5, x, 12, 17, 20, y, 9, 7, 4
Total frequency: 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100, giving x + y = 24. Since median is 525, the median class is 500-600. The cumulative frequency before the median class is 2 + 5 + x + 12 + 17 = 36 + x. Using the median formula: 525 = 500 + [(50 - (36 + x))/20] x 100. Simplifying: 25 = [(14 - x)/20] x 100, so 25 = 5(14 - x), giving 5 = 14 - x, hence x = 9. From x + y = 24, y = 15.
Find the mode of the following frequency distribution: Class: 25-35, 35-45, 45-55, 55-65, 65-75, 75-85 Frequency: 7, 31, 33, 17, 11, 1
The modal class is 45-55 (highest frequency = 33). Here l = 45, f1 = 33, f0 = 31, f2 = 17, h = 10. Applying the mode formula: Mode = 45 + [(33 - 31)/(66 - 31 - 17)] x 10 = 45 + [2/18] x 10 = 45 + 1.11 = 46.11 (approx). The modal value of the given distribution is approximately 46.11.
The mean of the following frequency distribution is 50. Find the missing frequencies f1 and f2. Class: 0-20, 20-40, 40-60, 60-80, 80-100 Frequency: 17, f1, 32, f2, 19 Total = 120
Total frequency: 17 + f1 + 32 + f2 + 19 = 120, giving f1 + f2 = 52. Mid-points are 10, 30, 50, 70, 90. Sum of fi x xi = 170 + 30f1 + 1600 + 70f2 + 1710 = 3480 + 30f1 + 70f2. Mean = 50, so 50 = (3480 + 30f1 + 70f2)/120. This gives 6000 = 3480 + 30f1 + 70f2, so 30f1 + 70f2 = 2520, which simplifies to 3f1 + 7f2 = 252. From f1 + f2 = 52, f1 = 52 - f2. Substituting: 3(52 - f2) + 7f2 = 252, giving 156 - 3f2 + 7f2 = 252, so 4f2 = 96, hence f2 = 24 and f1 = 28.
Find the median of the following data: Marks obtained: 20-30, 30-40, 40-50, 50-60, 60-70, 70-80 Number of students: 5, 10, 15, 20, 10, 5
Cumulative frequencies: 5, 15, 30, 50, 60, 65. Here n = 65 and n/2 = 32.5. The median class is 50-60 since cf = 50 is the first to exceed 32.5. Using the formula with l = 50, cf = 30, f = 20, h = 10: Median = 50 + [(32.5 - 30)/20] x 10 = 50 + [2.5/20] x 10 = 50 + 1.25 = 51.25. The median marks obtained by the students is 51.25.
If the mean of the following distribution is 27, find the value of p. Class: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 8, p, 12, 13, 10
Mid-points are 5, 15, 25, 35, 45. Sum of fi x xi = 40 + 15p + 300 + 455 + 450 = 1245 + 15p. Sum of fi = 43 + p. Given mean = 27, so 27 = (1245 + 15p)/(43 + p). Cross-multiplying: 27(43 + p) = 1245 + 15p, giving 1161 + 27p = 1245 + 15p. Simplifying: 12p = 84, hence p = 7.
The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp. Life time (hours): 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Frequency: 14, 56, 60, 86, 74, 62, 48
Cumulative frequencies: 14, 70, 130, 216, 290, 352, 400. Here n = 400, n/2 = 200. The median class is 3000-3500 since cf = 216 first exceeds 200. Using the formula with l = 3000, cf = 130, f = 86, h = 500: Median = 3000 + [(200 - 130)/86] x 500 = 3000 + [70/86] x 500 = 3000 + 406.98 = 3406.98 hours. The median lifetime is approximately 3407 hours.
Find the mode of the following distribution: Height (in cm): 150-155, 155-160, 160-165, 165-170, 170-175, 175-180 Number of women: 5, 8, 20, 12, 7, 3
The modal class is 160-165 (highest frequency = 20). Here l = 160, f1 = 20, f0 = 8, f2 = 12, h = 5. Mode = 160 + [(20 - 8)/(40 - 8 - 12)] x 5 = 160 + [12/20] x 5 = 160 + 3 = 163. The modal height of the women is exactly 163 cm.
The following distribution gives the daily wages of 50 workers of a factory. Draw a less than type ogive and find the median from the graph. Daily wages (Rs): 100-120, 120-140, 140-160, 160-180, 180-200 Frequency: 12, 14, 8, 6, 10
The less than cumulative frequencies are: Less than 120: 12, Less than 140: 26, Less than 160: 34, Less than 180: 40, Less than 200: 50. Plot the points (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) along with the starting point (100, 0). Join them with a smooth freehand curve. For the median, locate n/2 = 25 on y-axis, draw a horizontal line to the ogive, and drop a perpendicular to x-axis. The median from the graph is approximately Rs 138.57. This can be verified using the formula with median class 120-140.
If the median of the distribution given below is 28.5, find the values of x and y. Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Frequency: 5, x, 20, 15, y, 5 Total = 60
From total frequency: 5 + x + 20 + 15 + y + 5 = 60, so x + y = 15. Since the median is 28.5, the median class is 20-30. Cumulative frequency before the median class = 5 + x. Using the median formula: 28.5 = 20 + [(30 - (5 + x))/20] x 10. Simplifying: 8.5 = (25 - x)/2, giving 17 = 25 - x, so x = 8. From x + y = 15, we get y = 7. The missing frequencies are x = 8 and y = 7.
The following table gives the monthly consumption of electricity of 68 consumers. Find the median. Consumption (units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205 Frequency: 4, 5, 13, 20, 14, 8, 4
Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68. Here n = 68 and n/2 = 34. The median class is 125-145 since cf = 42 is the first to exceed 34. Using the formula: Median = 125 + [(34 - 22)/20] x 20 = 125 + [12/20] x 20 = 125 + 12 = 137 units. The median monthly electricity consumption is exactly 137 units.
The mode of the following data is 36. Find the missing frequency f. Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70 Frequency: 8, 10, f, 16, 12, 6, 7
Since mode is 36, the modal class is 30-40 (as 36 lies in this interval). Here l = 30, f1 = 16, f0 = f, f2 = 12, h = 10. Using the mode formula: 36 = 30 + [(16 - f)/(32 - f - 12)] x 10. Simplifying: 6 = [(16 - f)/(20 - f)] x 10, so 6(20 - f) = 10(16 - f). Expanding: 120 - 6f = 160 - 10f, giving 4f = 40, hence f = 10.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode. Consumption (units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205 Frequency: 4, 5, 13, 20, 14, 8, 4
Median: n = 68, n/2 = 34. Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68. Median class = 125-145. Median = 125 + [(34 - 22)/20] x 20 = 137 units. Mean (using a = 135, h = 20): Sum of fi x ui = -12-10-13+0+14+16+12 = 7. Mean = 135 + 20 x (7/68) = 137.06 units. Mode: Modal class = 125-145, Mode = 125 + [(20-13)/(40-13-14)] x 20 = 125 + [7/13] x 20 = 135.77 units. All three measures are close to 137, showing the distribution is nearly symmetric.
Draw a less than ogive and a more than ogive for the following data and hence find the median. Class: 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100 Frequency: 7, 5, 8, 10, 6, 6, 8
Less than cumulative frequencies: Less than 40: 7, Less than 50: 12, Less than 60: 20, Less than 70: 30, Less than 80: 36, Less than 90: 42, Less than 100: 50. More than cumulative frequencies: More than 30: 50, More than 40: 43, More than 50: 38, More than 60: 30, More than 70: 20, More than 80: 14, More than 90: 8. Plot both ogives on the same graph. The x-coordinate of their intersection point gives the median. From the graph, median is approximately 65 units. This can be verified using the formula: Median class 60-70, Median = 60 + [(25-20)/10] x 10 = 65.
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Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The mean of a grouped frequency distribution can be found using the direct method, assumed mean method, and step deviation method, and all three methods give the same result.
Reason (R): The assumed mean and step deviation methods are algebraic simplifications of the direct method formula, so they are mathematically equivalent.
Answer: (A)
Both Assertion and Reason are true, and the Reason correctly explains the Assertion. The step deviation method uses ui = (xi - a)/h, which when substituted back into the mean formula simplifies to the exact same expression as the direct method. All three methods are equivalent.
AR2
Assertion (A): The empirical relationship between mean, median, and mode is: 3 x Median = Mode + 2 x Mean.
Reason (R): This relationship holds exactly for all frequency distributions.
Answer: (C)
The Assertion is true - the empirical relationship 3 x Median = Mode + 2 x Mean is a well-known formula. However, the Reason is false because this relationship is only approximate and holds for moderately skewed distributions. For highly skewed distributions, the relationship may not hold. So the Assertion is true but the Reason is false.
AR3
Assertion (A): For the data: Class 10-20 (f=5), 20-30 (f=8), 30-40 (f=12), 40-50 (f=8), 50-60 (f=5), the median class is 30-40.
Reason (R): The median class is the class whose cumulative frequency is nearest to n/2.
Answer: (C)
The Assertion is true because n = 38, n/2 = 19. Cumulative frequencies are 5, 13, 25, 33, 38. The class 30-40 has cf = 25 which is the first to exceed 19, so it is the median class. The Reason is false because the median class is not the one nearest to n/2 but the one whose cumulative frequency first exceeds (or equals) n/2.
AR4
Assertion (A): The modal class is the class with the highest frequency.
Reason (R): Mode represents the value that occurs most frequently in a data set.
Answer: (A)
Both Assertion and Reason are true, and the Reason correctly explains the Assertion. In grouped data, the modal class is identified as the class interval having the maximum frequency. The mode formula is then applied to this class to find the exact mode value within the interval.
AR5
Assertion (A): The median of a grouped frequency distribution can be obtained graphically as the x-coordinate of the point of intersection of the less than ogive and the more than ogive.
Reason (R): At the point of intersection, the less than cumulative frequency equals the more than cumulative frequency, and both equal n/2.
Answer: (A)
Both Assertion and Reason are true, and the Reason correctly explains the Assertion. At the intersection point, the less than cf from below and the more than cf from above both reach n/2, which is precisely the definition of the median. This graphical method provides an elegant visual way to find the median.
AR6
Assertion (A): In the formula Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h, if f0 = f2, then the mode equals the mid-point of the modal class.
Reason (R): When f0 = f2, the formula simplifies to Mode = l + h/2, which is the mid-point of the modal class.
Answer: (A)
Both Assertion and Reason are true, and the Reason correctly explains the Assertion. When f0 = f2, Mode = l + [(f1 - f0)/(2f1 - 2f0)] x h = l + [1/2] x h = l + h/2, which is indeed the class mark (mid-point) of the modal class. The equal frequencies on both sides pull the mode to the centre of the class.
Case Study Questions
A school conducted a survey on the daily screen time (in minutes) of 50 students of Class 10. The data collected is shown below: Screen time (minutes): 0-30, 30-60, 60-90, 90-120, 120-150, 150-180 Number of students: 5, 8, 15, 12, 7, 3
(i) What is the modal class of the given data? (1 mark)
The modal class is the class with the highest frequency. Here, the class 60-90 has the highest frequency of 15. So the modal class is 60-90.
(ii) Find the mode of the data. (2 marks)
Modal class = 60-90, l = 60, f1 = 15, f0 = 8, f2 = 12, h = 30
Mode = 60 + [(15 - 8)/(30 - 8 - 12)] x 30
= 60 + [7/10] x 30
= 60 + 21
= 81 minutes
(iii) Find the median screen time of the students. (2 marks)
Cumulative frequencies: 5, 13, 28, 40, 47, 50
n = 50, n/2 = 25
Median class = 60-90 (cf = 28 first exceeds 25)
l = 60, cf = 13, f = 15, h = 30
Median = 60 + [(25 - 13)/15] x 30 = 60 + [12/15] x 30 = 60 + 24 = 84 minutes
A government agency conducted a survey on the literacy rates (in percentage) of 40 villages in a district. The data is summarised below: Literacy rate (%): 30-40, 40-50, 50-60, 60-70, 70-80, 80-90 Number of villages: 3, 5, 11, 10, 8, 3
(i) Find the mean literacy rate of the villages using the step deviation method. (2 marks)
Taking a = 55 and h = 10:
ui values: -2, -1, 0, 1, 2, 3
fi x ui: -6, -5, 0, 10, 16, 9
Sum of fi = 40, Sum of fi x ui = 24
Mean = 55 + 10 x (24/40) = 55 + 6 = 61%
(ii) Find the median literacy rate. (2 marks)
Cumulative frequencies: 3, 8, 19, 29, 37, 40
n = 40, n/2 = 20
Median class = 60-70 (cf = 29 first exceeds 20)
l = 60, cf = 19, f = 10, h = 10
Median = 60 + [(20 - 19)/10] x 10 = 60 + 1 = 61%
(iii) Use the empirical relationship to estimate the mode. (1 mark)
Using 3 x Median = Mode + 2 x Mean:
3 x 61 = Mode + 2 x 61
183 = Mode + 122
Mode = 183 - 122 = 61%
All three measures are equal at 61%, indicating a perfectly symmetric distribution.
MCQ Practice
MCQ1. The class mark of the class 10-20 is:
(A) 10
(B) 20
(C) 15
(D) 30
Answer: (C) - Class mark = (lower limit + upper limit)/2 = (10 + 20)/2 = 15.
MCQ2. Construction of a cumulative frequency table is useful in determining the:
(A) Mean
(B) Median
(C) Mode
(D) All of the above
Answer: (B) - The cumulative frequency table is used specifically to identify the median class and apply the median formula. Mean and mode do not require cumulative frequencies.
MCQ3. The empirical relationship between the three measures of central tendency is:
(A) 3 x Mean = Mode + 2 x Median
(B) 3 x Median = Mode + 2 x Mean
(C) 3 x Mode = Mean + 2 x Median
(D) Mean + Median + Mode = 3
Answer: (B) - The empirical relationship is 3 x Median = Mode + 2 x Mean. This can also be written as Mode = 3 x Median - 2 x Mean.
MCQ4. If the mean of a frequency distribution is 8.1 and the sum of fi x xi = 132 + 5k, and sum of fi = 20, then the value of k is:
(A) 3
(B) 4
(C) 5
(D) 6
Answer: (D) - Mean = (Sum of fi x xi)/(Sum of fi). So 8.1 = (132 + 5k)/20. Therefore 162 = 132 + 5k, giving 5k = 30 and k = 6.
MCQ5. In the formula for mode of grouped data, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h, what does f0 represent?
(A) Frequency of the modal class
(B) Frequency of the class succeeding the modal class
(C) Frequency of the class preceding the modal class
(D) Total frequency
Answer: (C) - In the mode formula, f0 is the frequency of the class just before (preceding) the modal class. f1 is the frequency of the modal class itself, and f2 is the frequency of the class just after (succeeding) the modal class.
MCQ6. The median of a frequency distribution is found graphically with the help of:
(A) Histogram
(B) Frequency polygon
(C) Ogive
(D) Bar graph
Answer: (C) - An ogive (cumulative frequency curve) is used to find the median graphically. The median corresponds to the x-value where the cumulative frequency equals n/2.
MCQ7. If the mode of a data is 45 and the mean is 27, then the median is:
(A) 30
(B) 33
(C) 36
(D) 39
Answer: (B) - Using the empirical relationship: 3 x Median = Mode + 2 x Mean = 45 + 54 = 99. So Median = 99/3 = 33.
MCQ8. While computing the mean of grouped data, we assume that the frequencies are:
(A) Evenly distributed over all the classes
(B) Centred at the class marks of the classes
(C) Centred at the upper limits of the classes
(D) Centred at the lower limits of the classes
Answer: (B) - When computing the mean of grouped data, we use class marks (mid-points) as representative values. This assumes all observations in a class are centred at the class mark.
MCQ9. If xi's are the mid-points of the class intervals of grouped data, fi's are the corresponding frequencies and x-bar is the mean, then sum of fi x (xi - x-bar) equals:
(A) 0
(B) -1
(C) 1
(D) 2
Answer: (A) - Sum of fi x (xi - x-bar) = Sum of fi x xi - x-bar x Sum of fi = Sum of fi x xi - (Sum of fi x xi / Sum of fi) x Sum of fi = Sum of fi x xi - Sum of fi x xi = 0. This is a fundamental property of the arithmetic mean.
MCQ10. The abscissa of the point of intersection of the less than type ogive and the more than type ogive gives:
(A) Mean
(B) Median
(C) Mode
(D) Sum of frequencies
Answer: (B) - The x-coordinate (abscissa) of the intersection point of the less than ogive and the more than ogive gives the median of the data. At this point, the cumulative frequency from both directions equals n/2.
MCQ11. For the following distribution: Class 0-5 (f=10), 5-10 (f=15), 10-15 (f=12), 15-20 (f=20), 20-25 (f=9), the sum of lower limits of the median class and the modal class is:
(A) 15
(B) 25
(C) 30
(D) 35
Answer: (B) - n = 66, n/2 = 33. Cumulative frequencies: 10, 25, 37, 57, 66. Median class = 10-15 (cf = 37 is the first to exceed 33, lower limit = 10). Modal class = 15-20 (highest frequency = 20, lower limit = 15). Sum of lower limits = 10 + 15 = 25.
MCQ12. Consider the following frequency distribution: Class 12-17 (f=2), 17-22 (f=5), 22-27 (f=8), 27-32 (f=12), 32-37 (f=3). The upper limit of the median class is:
(A) 17
(B) 22
(C) 27
(D) 32
Answer: (C) - n = 30, n/2 = 15. Cumulative frequencies: 2, 7, 15, 27, 30. The median class is 22-27 (cf = 15 is the first value that equals or exceeds n/2 = 15). The upper limit of this median class is 27.
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Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Write the formula for finding the mode of grouped data.
Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h, where l is the lower limit of the modal class, f1 is the frequency of the modal class, f0 is the frequency of the preceding class, f2 is the frequency of the succeeding class, and h is the class size.
If the mean and mode of a data are 24 and 21 respectively, find the median.
Using 3 x Median = Mode + 2 x Mean: 3 x Median = 21 + 48 = 69. Median = 69/3 = 23.
What is the class mark of the class interval 15.5-20.5?
Class mark = (15.5 + 20.5)/2 = 36/2 = 18.
For a data with mode = 20 and mean = 17, find the median using the empirical relation.
3 x Median = Mode + 2 x Mean = 20 + 34 = 54. Median = 54/3 = 18.
In an ogive, the x-coordinate of the point at which the less than ogive and more than ogive intersect gives which measure of central tendency?
The x-coordinate of the intersection point of the two ogives gives the median of the data.
2-3 Mark Questions
Find the mode of the following data: Class: 0-20, 20-40, 40-60, 60-80, 80-100 Frequency: 10, 8, 12, 16, 4
Modal class = 60-80 (highest frequency = 16)
l = 60, f1 = 16, f0 = 12, f2 = 4, h = 20
Mode = 60 + [(16 - 12)/(32 - 12 - 4)] x 20
= 60 + [4/16] x 20
= 60 + 5
= 65
Find the mean of the following distribution using step deviation method: Class: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 6, 10, 12, 8, 4
Taking a = 25, h = 10: ui values are -2, -1, 0, 1, 2.
fi x ui: -12, -10, 0, 8, 8.
Sum of fi = 40, Sum of fi x ui = -6.
Mean = 25 + 10 x (-6/40) = 25 - 1.5 = 23.5
The mean of the following distribution is 50. Find the missing frequency p. Class: 0-20, 20-40, 40-60, 60-80, 80-100 Frequency: 7, p, 10, 9, 13
Mid-points: 10, 30, 50, 70, 90. Sum of fi x xi = 70 + 30p + 500 + 630 + 1170 = 2370 + 30p. Sum of fi = 39 + p.
Mean = 50: 50 = (2370 + 30p)/(39 + p). Cross-multiply: 1950 + 50p = 2370 + 30p. So 20p = 420, p = 21.
Find the median of the following data: Class: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 5, 8, 20, 7, 5
cf: 5, 13, 33, 40, 45. n = 45, n/2 = 22.5.
Median class = 20-30 (cf = 33 first exceeds 22.5).
l = 20, cf = 13, f = 20, h = 10.
Median = 20 + [(22.5 - 13)/20] x 10 = 20 + 4.75 = 24.75
3-4 Mark Questions
The following distribution gives the daily income of 50 workers. If the median income is Rs 138.57, find the missing frequencies a and b. Income (Rs): 100-120, 120-140, 140-160, 160-180, 180-200 Frequency: 12, a, b, 6, 10
Total: 12 + a + b + 6 + 10 = 50, so a + b = 22 ... (i)
Median = 138.57, Median class = 120-140.
cf before median class = 12, f = a, n/2 = 25.
138.57 = 120 + [(25 - 12)/a] x 20
18.57 = 260/a
a = 260/18.57 = 14 (approx.)
From (i): b = 22 - 14 = 8.
So a = 14, b = 8.
Find the mean, median, and mode of the following data: Class: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 8, 16, 36, 34, 6
Mean (a = 25, h = 10): fi x ui: -16, -16, 0, 34, 12. Sum = 14. Mean = 25 + 10 x (14/100) = 25 + 1.4 = 26.4.
Median: cf: 8, 24, 60, 94, 100. n = 100, n/2 = 50. Median class = 20-30. Median = 20 + [(50 - 24)/36] x 10 = 20 + 7.22 = 27.22.
Mode: Modal class = 20-30 (f = 36). Mode = 20 + [(36 - 16)/(72 - 16 - 34)] x 10 = 20 + [20/22] x 10 = 20 + 9.09 = 29.09.
Verification: 3 x Median = 81.66. Mode + 2 x Mean = 29.09 + 52.8 = 81.89. The values are approximately equal.
Draw both ogives for the following data and find the median from the graph. Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Frequency: 5, 8, 12, 14, 8, 3
Less than cf: (10,5), (20,13), (30,25), (40,39), (50,47), (60,50).
More than cf: (0,50), (10,45), (20,37), (30,25), (40,11), (50,3).
Plot both curves. They intersect at x = 30, y = 25.
Median from graph = 30.
Verification: n = 50, n/2 = 25. Median class = 30-40. Median = 30 + [(25-25)/14] x 10 = 30 + 0 = 30. The graphical and formula values match exactly.
Key Concepts
Important Formulas
Study Plan - Statistics
Mean of Grouped Data - All Three Methods
Read NCERT theory on direct method, assumed mean method, and step deviation method. Understand when to use each method. Solve Exercise 13.1 Q1-Q5. Pay special attention to Q5 (missing frequency) and Q4 (all three methods on same data).
Mean - Remaining Questions and Discontinuous Classes
Complete Exercise 13.1 Q6-Q9. Focus on Q7 which requires converting discontinuous classes to continuous. Practice choosing the best assumed mean. Revise step deviation shortcuts.
Mode of Grouped Data
Learn the mode formula and its components (l, f1, f0, f2, h). Solve all 6 questions of Exercise 13.2. Practice identifying the modal class quickly. Note how Q1 and Q3 ask for both mode and mean.
Median of Grouped Data
Learn to build cumulative frequency tables and apply the median formula. Solve Exercise 13.3 Q1-Q4. Focus on Q2 (find missing frequencies given median) and Q4 (converting discontinuous to continuous classes).
Median - Remaining Questions and Ogive
Complete Exercise 13.3 Q5-Q7. Then solve all 3 questions of Exercise 13.4. Practice drawing less than and more than ogives. Learn to read the median from the graph.
PYQs, MCQs, Assertion-Reason, and Case Studies
Solve all 15 previous year questions from this page. Practice the 12 MCQs and 6 assertion-reason questions. Attempt both case study questions under timed conditions (15 minutes each). Revise the empirical relationship.
Revision and Important Questions
Revise all formulas and the common mistakes list. Solve all important questions (1-mark, 2-3 mark, 3-4 mark). Time yourself on the 3-4 mark questions (6-8 minutes each). Review the chapter summary for quick revision before exams.
Common Mistakes to Avoid
- ✗Using the total frequency n instead of n/2 when finding the median class from the cumulative frequency table
- ✗In the median formula, using the cumulative frequency of the median class itself instead of the cumulative frequency of the class just before (preceding) the median class
- ✗In the mode formula, confusing f0 (preceding class frequency) and f2 (succeeding class frequency) - swapping them gives a wrong answer
- ✗Forgetting to convert discontinuous class intervals (like 1-10, 11-20) to continuous form (0.5-10.5, 10.5-20.5) before applying the median or mode formula
- ✗Using the wrong class mark (mid-point) when the class intervals are not equal or when they have been converted from discontinuous to continuous form
- ✗In the step deviation method, forgetting to multiply back by h in the final step: Mean = a + h x (Sum of fi x ui / Sum of fi), not just a + (Sum of fi x ui / Sum of fi)
- ✗Drawing the ogive by plotting frequencies instead of cumulative frequencies on the y-axis
- ✗In less than ogive, plotting lower class limits instead of upper class limits (and vice versa for more than ogive)
- ✗Arithmetic errors in computing Sum of fi x di or Sum of fi x ui, especially when dealing with negative values - always double-check the column totals
- ✗Applying the empirical relation Mode = 3 Median - 2 Mean as an exact formula rather than an approximation, leading to answers that do not match when verified
Tips for Board Exam
- ✓Memorise the three formulas: Mean (step deviation) = a + h x (Sum of fi x ui / Sum of fi), Mode = l + [(f1-f0)/(2f1-f0-f2)] x h, Median = l + [(n/2-cf)/f] x h. These three formulas cover almost all questions in this chapter.
- ✓For the mean, always use the step deviation method in exams. It reduces the numbers to small integers (ui values like -2, -1, 0, 1, 2), minimising calculation errors. Choose the assumed mean 'a' as the class mark of the class with the highest frequency.
- ✓When finding the median class, build the cumulative frequency column carefully. The median class is the first class whose cf is greater than or equal to n/2. Mark it clearly before applying the formula.
- ✓In missing frequency problems (find x given mean/median), set up the equation carefully. These problems are worth 3-5 marks and follow a predictable pattern: use the given measure of central tendency to create a linear equation in the unknown.
- ✓For ogive questions, always start by making the cumulative frequency table first. For less than ogive, plot (upper class boundary, cf). For more than ogive, plot (lower class boundary, n - cf from above). Add the starting/ending point for a complete curve.
- ✓The empirical relationship 3 x Median = Mode + 2 x Mean is useful for 1-mark MCQs. If you know any two measures, you can estimate the third. Remember this is an approximation, not an exact formula.
- ✓When the question asks to 'compare and interpret' mean and mode, always write a one-line interpretation linking the statistical measure to the real-life context (e.g., 'the mode of 36.82 years means maximum patients are in the 35-45 age group').
- ✓In the board exam, show all steps: write the frequency table, show the formula with substitution, and clearly box/bold the final answer. Marks are given for the method even if the final calculation has a small error.
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Start Statistics - Choose Your ModeAbout Statistics - Class 10 Mathematics
Chapter 13 Statistics is part of the Statistics and Probability unit in CBSE Class 10 Mathematics and carries 5-6 marks in the board exam 2027. This chapter focuses entirely on three measures of central tendency for grouped data: mean, mode, and median. It also introduces cumulative frequency distributions and their graphical representation as ogives. This is one of the most scoring chapters in the entire syllabus because every question follows a fixed formula-based pattern with no proofs or abstract reasoning.
What This Chapter Covers
The chapter teaches students to calculate the mean, mode, and median of grouped (classified) data using specific formulas. It also covers the graphical representation of cumulative frequency distributions through ogives, which can be used to find the median visually. The chapter builds on the concepts of mean, median, and mode for ungrouped data studied in earlier classes and extends them to frequency distributions.
Mean of Grouped Data
The mean (average) of grouped data can be calculated using three methods, all of which give the same answer but differ in computational complexity.
The Direct Method uses the formula Mean = (Sum of fi x xi) / (Sum of fi), where xi is the class mark (midpoint) of each class and fi is the frequency. This method works well when the values of fi and xi are small, but becomes cumbersome when dealing with large numbers.
The Assumed Mean Method simplifies calculations by choosing a convenient value 'a' (usually the class mark of the middle class or the class with the highest frequency) and computing deviations di = xi - a. The formula becomes Mean = a + (Sum of fi x di) / (Sum of fi). This reduces the size of the numbers being multiplied.
The Step Deviation Method goes one step further by dividing the deviations by the class size h, giving ui = (xi - a) / h. The formula becomes Mean = a + h x (Sum of fi x ui) / (Sum of fi). Since ui values are typically small integers like -3, -2, -1, 0, 1, 2, 3, this method is the fastest and least error-prone. It is the recommended method for board exams.
All three methods are algebraically equivalent. The assumed mean and step deviation methods are simply mathematical rearrangements of the direct method formula. They do not introduce any approximation.
Mode of Grouped Data
The mode is the value that occurs most frequently. For grouped data, we first identify the modal class, which is the class interval with the highest frequency. The mode is then calculated using the formula:
Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] x h
Here, l is the lower limit of the modal class, f1 is the frequency of the modal class, f0 is the frequency of the class just before (preceding) the modal class, f2 is the frequency of the class just after (succeeding) the modal class, and h is the class size.
The formula essentially interpolates within the modal class based on how much higher its frequency is compared to its neighbouring classes. If f0 equals f2, the mode falls exactly at the midpoint of the modal class. If f0 is larger than f2, the mode shifts towards the lower end, and vice versa.
It is important to note that this formula applies only when the class intervals are of equal width. For unequal class intervals, the concept of modal class needs to be handled differently, but such cases are not asked at the Class 10 level.
Median of Grouped Data
The median is the middle value that divides the data into two equal halves. For grouped data, we use cumulative frequencies to locate the median class and then apply the formula:
Median = l + [(n/2 - cf) / f] x h
Here, l is the lower limit of the median class, n is the total frequency, 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.
To find the median class, compute n/2 and look at the cumulative frequency column. The first class whose cumulative frequency equals or exceeds n/2 is the median class.
The most common error in applying this formula is using the wrong value of cf. It must be the cumulative frequency of the class just before the median class, not the cumulative frequency of the median class itself.
Cumulative Frequency and Ogives
A cumulative frequency distribution can be of two types. The less than type records how many observations fall below each upper class boundary. The more than type records how many observations are above each lower class boundary.
When these cumulative frequencies are plotted against the corresponding class boundaries and joined by a smooth curve, the result is called an ogive (pronounced oh-jive).
The less than ogive is an increasing S-shaped curve, plotted using (upper class boundary, cumulative frequency). The more than ogive is a decreasing curve, plotted using (lower class boundary, cumulative frequency from total downwards).
The median can be found graphically by locating n/2 on the y-axis of either ogive and reading the corresponding x-value. Alternatively, if both ogives are drawn on the same graph, their point of intersection has the median as its x-coordinate.
Empirical Relationship Between Mean, Median, and Mode
For moderately skewed distributions, there is an approximate empirical relationship: 3 x Median = Mode + 2 x Mean. This can be rewritten as Mode = 3 x Median - 2 x Mean. This relationship is useful for quick estimation when two of the three measures are known. However, students should remember that this is an approximation and may not hold exactly for all data sets.
CBSE 2026-27 Syllabus Details
For the CBSE 2026-27 session, Chapter 13 Statistics includes all 4 exercises: Exercise 13.1 (9 questions on mean using direct, assumed mean, and step deviation methods), Exercise 13.2 (6 questions on mode), Exercise 13.3 (7 questions on median), and Exercise 13.4 (3 questions on ogives). The total is 25 questions. No exercises have been removed from this chapter.
How This Chapter Is Tested in Board Exams
Statistics is part of the Statistics and Probability unit which carries 11 marks total (shared with Chapter 14 Probability). From Chapter 13 alone, students can typically expect 5-6 marks distributed as follows. One MCQ worth 1 mark tests a basic concept (class mark, empirical relation, identifying median/modal class). One short answer worth 3 marks asks students to find the mean, mode, or median of a given distribution using the standard formula. One longer question worth 3-5 marks may ask for missing frequencies given the mean or median, or may combine two measures (find mean and mode, then compare).
Ogive questions (Exercise 13.4) appear less frequently in the main exam but are commonly asked in case study questions and internal assessments. When asked, they typically carry 4-5 marks and require drawing the ogive and finding the median graphically.
Recent CBSE trends show increased emphasis on missing frequency problems (given mean = X, find the missing frequency f) and combined questions (find median and mode, then verify using the empirical relation). Case study questions on census data, school surveys, and environmental data have appeared since 2023.
Connection to Other Chapters
Chapter 13 connects directly to Chapter 14 (Probability) as both are in the same unit. The concept of frequency distributions in Chapter 13 provides the foundation for understanding probability distributions in higher classes. The cumulative frequency concept from this chapter is also related to the cumulative distribution function (CDF) studied in Class 11 and 12 Mathematics.
Students preparing for competitive exams (NTSE, Olympiad) should note that Statistics at the Class 10 level is limited to measures of central tendency for grouped data. Measures of dispersion (range, standard deviation, variance) are not covered and will be introduced in Class 11. The ogive concepts, however, provide a useful visual foundation for understanding distribution shapes and quartiles.
Key Differences Between Mean, Median, and Mode
The mean is affected by extreme values (outliers). If a data set has a few very large or very small values, the mean gets pulled towards those extremes. The median is resistant to outliers because it depends only on the middle value. The mode identifies the most common value and is useful for categorical data or when we want to know the most typical value. In symmetric distributions, all three measures are approximately equal. In skewed distributions, they differ, and the choice of which measure to use depends on the context.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 13 for CBSE 2026-27?
Chapter 13 Statistics has 4 exercises in the CBSE 2026-27 syllabus: Exercise 13.1 (9 questions on mean of grouped data using direct, assumed mean, and step deviation methods), Exercise 13.2 (6 questions on mode of grouped data), Exercise 13.3 (7 questions on median of grouped data), and Exercise 13.4 (3 questions on ogives). All 4 exercises are part of the board exam syllabus.
What are the three methods to find the mean of grouped data?
The three methods are: (1) Direct Method: Mean = (Sum of fi x xi)/(Sum of fi). (2) Assumed Mean Method: Mean = a + (Sum of fi x di)/(Sum of fi), where di = xi - a. (3) Step Deviation Method: Mean = a + h x (Sum of fi x ui)/(Sum of fi), where ui = (xi - a)/h. All three give the same answer. The step deviation method is recommended for board exams as it simplifies calculations.
How many marks does Chapter 13 Statistics carry in the CBSE Class 10 board exam?
Chapter 13 is part of the Statistics and Probability unit which carries 11 marks total. From Chapter 13 alone, students can expect 5-6 marks: typically 1 MCQ (1 mark) + 1 short answer on mean/mode/median (3 marks) + possibly 1 longer question on missing frequencies or ogive (3-5 marks).
What is the difference between a less than ogive and a more than ogive?
A less than ogive plots upper class boundaries on the x-axis against cumulative frequencies (running total from the beginning) on the y-axis, giving an increasing S-shaped curve. A more than ogive plots lower class boundaries on the x-axis against cumulative frequencies counted from the total downwards, giving a decreasing curve. Both can be used to find the median graphically.
What is the empirical relationship between mean, median, and mode?
The empirical relationship is: 3 x Median = Mode + 2 x Mean. This can also be written as Mode = 3 x Median - 2 x Mean. This relationship is approximate and works well for moderately skewed distributions. It is useful for 1-mark MCQs where you need to estimate one measure from the other two.
How to find the median from an ogive?
Method 1: Draw a less than ogive. Locate n/2 on the y-axis, draw a horizontal line to the curve, then drop a vertical line to the x-axis. The x-value is the median. Method 2: Draw both less than and more than ogives on the same axes. Their point of intersection gives the median as the x-coordinate.
When should I use assumed mean method vs step deviation method?
Use the step deviation method when all class intervals are equal (same class size h). It reduces calculations to small integers. Use the assumed mean method when class intervals are unequal. In practice, for board exams, the step deviation method is almost always the best choice since NCERT questions use equal class intervals.
What type of questions from Statistics are most frequently asked in CBSE board exams?
The most frequently asked types are: (1) Find the mean using step deviation method (3 marks). (2) Find missing frequency given the mean or median (3-5 marks). (3) Find the mode of grouped data (3 marks). (4) Find the median of grouped data (3 marks). (5) MCQ on empirical relation or identifying median/modal class (1 mark). Ogive drawing questions appear less frequently but are common in case study format.