Class 10 · Mathematics

Chapter 13: Statistics

Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.

13.1–13.2Mean of grouped dataDirect, assumed mean and step-deviation

Class mark

For grouped data we assume each class is centred at its class mark:

xᵢ = lower limit + upper limit2

Three methods

Direct: x̄ = ΣfᵢxᵢΣfᵢAssumed mean: x̄ = a + ΣfᵢdᵢΣfᵢ, dᵢ = xᵢ − aStep deviation: x̄ = a + (ΣfᵢuᵢΣfᵢ) × h, uᵢ = xᵢ − ah

All three give the same mean. The last two only make the arithmetic easier.
Choose a (the assumed mean) as a class mark near the middle of the table.
Use the direct method when xᵢ and fᵢ are small; step deviation when the class marks are large and the class sizes equal.
With unequal class sizes, the step-deviation method still works if h divides all the dᵢ.

Example

Daily wages of 50 workers (Ex 13.1, Q2), step deviation with a = 550, h = 20:

Step-deviation method with a = 550, h = 20, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	50			−12

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 550 + (−1250) × 20 = 545.2

Finding a missing frequency

Write Σfᵢ and Σfᵢxᵢ in terms of the unknown f, set Σfᵢxᵢ / Σfᵢ equal to the given mean, and solve.

13.3Mode of grouped dataModal class and the formula

Modal class

The mode is the value that occurs most often. For grouped data, first find the modal class, the class with the highest frequency.

The mode lies in the class with the highest frequency, pulled towards the taller neighbour.

Formula

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h

l = lower limit of the modal class, h = class size
f₁ = frequency of the modal class, f₀ = of the class before it, f₂ = of the class after it
If the modal class is the first class, f₀ = 0; if it is the last, f₂ = 0.

When to use which

Mean: when every value matters (average marks, average income).
Mode: the most popular value (most common shoe size, most frequent number of cars).
Median: when extreme values would distort the mean (house prices, incomes).

13.4Median of grouped dataCumulative frequency and the formula

Cumulative frequency

The cumulative frequency (cf) of a class is the sum of the frequencies of all classes up to and including it. The total is n.

Cumulative frequencies of the electricity data. The median class is 125 – 145.

Formula

Median = l + (n/2 − cff) × h

Median class: the first class whose cumulative frequency is greater than n/2.
l = lower limit of the median class, cf = cumulative frequency of the class before it,
f = frequency of the median class, h = class size.

Continuous classes

The formulas for the mode and the median need continuous classes. Classes like 118 – 126, 127 – 135 have a gap of 1: subtract 0.5 from each lower limit and add 0.5 to each upper limit to get 117.5 – 126.5, 126.5 – 135.5, …

'Less than' or 'below' tables already give cumulative frequencies; subtract consecutive values to get the class frequencies.

Empirical relation

3 Median = Mode + 2 Mean

This relation holds approximately for moderately skewed data and can be used to estimate one measure from the other two.

RefQuick referenceFormulas and slips

Formulas

Measure	Formula
Mean (direct)	Σfᵢxᵢ / Σfᵢ
Mean (assumed)	a + Σfᵢdᵢ / Σfᵢ
Mean (step deviation)	a + (Σfᵢuᵢ / Σfᵢ) × h
Mode	l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
Median	l + [(n/2 − cf)/f] × h
Empirical	3 Median = Mode + 2 Mean

Common slips

Using the cf of the median class instead of the class before it.
Choosing the median class from the frequencies instead of the cumulative frequencies.
Using f₀ and f₂ from the wrong classes.
Applying the formulas to discontinuous classes without adjusting.
Forgetting to multiply by h in the step-deviation method.

Try each problem first, then open the solution to check your working.

13.2MeanChoose a method · 5 problems

Q1Direct method

Find the mean of the marks.

Marks	0 − 10	10 − 20	20 − 30	30 − 40	40 − 50
Students	4	6	10	12	8

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Class	fᵢ	xᵢ	fᵢxᵢ
0 − 10	4	5	20
10 − 20	6	15	90
20 − 30	10	25	250
30 − 40	12	35	420
40 − 50	8	45	360
Total	40		1140

x̄ = ΣfᵢxᵢΣfᵢ = 114040 = 28.5

Answer: 28.5

Q2Step deviation

Find the mean daily earning.

Earning (₹)	100 − 120	120 − 140	140 − 160	160 − 180	180 − 200
Workers	12	14	8	6	10

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Step-deviation method with a = 150, h = 20, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
100 − 120	12	110	−2	−24
120 − 140	14	130	−1	−14
140 − 160	8	150	0	0
160 − 180	6	170	1	6
180 − 200	10	190	2	20
Total	50			−12

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 150 + (−1250) × 20 = 145.2

Answer: ₹145.2

Q3Mean of heights

Find the mean height of the students.

Height (cm)	150 − 155	155 − 160	160 − 165	165 − 170	170 − 175
Students	7	12	18	8	5

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Step-deviation method with a = 162.5, h = 5, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
150 − 155	7	152.5	−2	−14
155 − 160	12	157.5	−1	−12
160 − 165	18	162.5	0	0
165 − 170	8	167.5	1	8
170 − 175	5	172.5	2	10
Total	50			−8

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 162.5 + (−850) × 5 = 161.7

Answer: 161.7 cm

Q4Assumed mean

Find the mean using the assumed-mean method.

Class	10 − 20	20 − 30	30 − 40	40 − 50	50 − 60	60 − 70	70 − 80
Frequency	2	3	5	7	4	3	1

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Take the assumed mean a = 45, dᵢ = xᵢ − a.

Class	fᵢ	xᵢ	dᵢ	fᵢdᵢ
10 − 20	2	15	−30	−60
20 − 30	3	25	−20	−60
30 − 40	5	35	−10	−50
40 − 50	7	45	0	0
50 − 60	4	55	10	40
60 − 70	3	65	20	60
70 − 80	1	75	30	30
Total	25			−40

x̄ = a + ΣfᵢdᵢΣfᵢ = 45 + −4025 = 43.4

Answer: 43.4

Q5Missing frequency

The mean of the following distribution is 26. Find p.

Class	0 − 10	10 − 20	20 − 30	30 − 40	40 − 50
Frequency	5	8	p	12	5

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Given: class marks 5, 15, 25, 35, 45

Use: Σfᵢxᵢ = mean × Σfᵢ

Σfᵢ = 30 + p; Σfᵢxᵢ = 25 + 120 + 25p + 420 + 225 = 790 + 25p

(790 + 25p)/(30 + p) = 26 ⇒ 790 + 25p = 780 + 26p

Answer: p = 10

13.3ModeModal class first · 4 problems

Q6Mode

Find the mode of the distribution.

Class	0 − 20	20 − 40	40 − 60	60 − 80	80 − 100	100 − 120	120 − 140
Frequency	6	8	10	12	6	5	3

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The highest frequency is 12, so the modal class is 60 − 80. l = 60, h = 20, f₁ = 12, f₀ = 10, f₂ = 6.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 60 + (12 − 1024 − 10 − 6) × 20 = 65

Answer: 65

Q7Mode of heights

Find the modal height.

Height (cm)	150 − 155	155 − 160	160 − 165	165 − 170	170 − 175
Students	7	12	18	8	5

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The highest frequency is 18, so the modal class is 160 − 165. l = 160, h = 5, f₁ = 18, f₀ = 12, f₂ = 8.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 160 + (18 − 1236 − 12 − 8) × 5 = 161.875

Answer: 161.875 cm

Q8Mode of earnings

Find the modal daily earning.

Earning (₹)	100 − 120	120 − 140	140 − 160	160 − 180	180 − 200
Workers	12	14	8	6	10

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The highest frequency is 14, so the modal class is 120 − 140. l = 120, h = 20, f₁ = 14, f₀ = 12, f₂ = 8.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 120 + (14 − 1228 − 12 − 8) × 20 = 125

Answer: ₹125

Q9Mode of small data

Find the mode.

Class	1 − 3	3 − 5	5 − 7	7 − 9	9 − 11
Frequency	6	9	15	9	1

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The highest frequency is 15, so the modal class is 5 − 7. l = 5, h = 2, f₁ = 15, f₀ = 9, f₂ = 9.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 5 + (15 − 930 − 9 − 9) × 2 = 6

Answer: 6

13.4MedianCumulative frequency first · 7 problems

Q10Median

Find the median.

Class	10 − 20	20 − 30	30 − 40	40 − 50	50 − 60	60 − 70	70 − 80
Frequency	2	3	5	7	4	3	1

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Class	Frequency	Cumulative frequency
10 − 20	2	2
20 − 30	3	5
30 − 40	5	10
40 − 50	7	17
50 − 60	4	21
60 − 70	3	24
70 − 80	1	25

n = 25, n/2 = 12.5. The first cumulative frequency greater than 12.5 is 17, so the median class is 40 − 50: l = 40, cf = 10, f = 7, h = 10.

Median = l + (n/2 − cff) × h = 40 + (12.5 − 107) × 10 = 43.57

Answer: 43.57

Q11Median of heights

Find the median height.

Height (cm)	150 − 155	155 − 160	160 − 165	165 − 170	170 − 175
Students	7	12	18	8	5

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Class	Frequency	Cumulative frequency
150 − 155	7	7
155 − 160	12	19
160 − 165	18	37
165 − 170	8	45
170 − 175	5	50

n = 50, n/2 = 25. The first cumulative frequency greater than 25 is 37, so the median class is 160 − 165: l = 160, cf = 19, f = 18, h = 5.

Median = l + (n/2 − cff) × h = 160 + (25 − 1918) × 5 = 161.67

Answer: 161.67 cm

Q12All three

Find the mean, median and mode.

Class	0 − 20	20 − 40	40 − 60	60 − 80	80 − 100	100 − 120	120 − 140
Frequency	6	8	10	12	6	5	3

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Step-deviation method with a = 70, h = 20, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
0 − 20	6	10	−3	−18
20 − 40	8	30	−2	−16
40 − 60	10	50	−1	−10
60 − 80	12	70	0	0
80 − 100	6	90	1	6
100 − 120	5	110	2	10
120 − 140	3	130	3	9
Total	50			−19

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 70 + (−1950) × 20 = 62.4

Class	Frequency	Cumulative frequency
0 − 20	6	6
20 − 40	8	14
40 − 60	10	24
60 − 80	12	36
80 − 100	6	42
100 − 120	5	47
120 − 140	3	50

n = 50, n/2 = 25. The first cumulative frequency greater than 25 is 36, so the median class is 60 − 80: l = 60, cf = 24, f = 12, h = 20.

Median = l + (n/2 − cff) × h = 60 + (25 − 2412) × 20 = 61.67

The highest frequency is 12, so the modal class is 60 − 80. l = 60, h = 20, f₁ = 12, f₀ = 10, f₂ = 6.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 60 + (12 − 1024 − 10 − 6) × 20 = 65

Answer: mean 62.4, median 61.67, mode 65

Q13Median of earnings

Find the median daily earning.

Earning (₹)	100 − 120	120 − 140	140 − 160	160 − 180	180 − 200
Workers	12	14	8	6	10

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Class	Frequency	Cumulative frequency
100 − 120	12	12
120 − 140	14	26
140 − 160	8	34
160 − 180	6	40
180 − 200	10	50

n = 50, n/2 = 25. The first cumulative frequency greater than 25 is 26, so the median class is 120 − 140: l = 120, cf = 12, f = 14, h = 20.

Median = l + (n/2 − cff) × h = 120 + (25 − 1214) × 20 = 138.57

Answer: ₹138.57

Q14Median of marks

Find the median mark.

Marks	0 − 10	10 − 20	20 − 30	30 − 40	40 − 50
Students	4	6	10	12	8

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Class	Frequency	Cumulative frequency
0 − 10	4	4
10 − 20	6	10
20 − 30	10	20
30 − 40	12	32
40 − 50	8	40

n = 40, n/2 = 20. The first cumulative frequency greater than 20 is 32, so the median class is 30 − 40: l = 30, cf = 20, f = 12, h = 10.

Median = l + (n/2 − cff) × h = 30 + (20 − 2012) × 10 = 30

Answer: 30

Q15Empirical relation

For a distribution, the mode is 24.5 and the mean is 29.75. Estimate the median.

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Use: 3 Median = Mode + 2 Mean

3 Median = 24.5 + 59.5 = 84

Answer: 28

Q16Less-than table

Convert into a frequency table and find the median: less than 10: 5, less than 20: 9, less than 30: 17, less than 40: 29, less than 50: 45.

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Given: frequencies 5, 4, 8, 12, 16 for 0–10, …, 40–50; n = 45

n/2 = 22.5; cumulative frequencies 5, 9, 17, 29, 45: median class 30 − 40

Median = 30 + (22.5 − 17)/12 × 10 = 30 + 4.58

Answer: ≈ 34.58

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

Ex 13.1Exercise 13.1Mean of grouped data · 9 questions

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. Which method did you use for finding the mean, and why?

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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

Class	fᵢ	xᵢ	fᵢxᵢ
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
Total	20		162

x̄ = ΣfᵢxᵢΣfᵢ = 16220 = 8.1

The direct method is used because the class marks and frequencies are small numbers.

Answer: 8.1 plants per house (direct method)

Q2Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

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Daily wages (₹)	500 − 520	520 − 540	540 − 560	560 − 580	580 − 600
Number of workers	12	14	8	6	10

Step-deviation method with a = 550, h = 20, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	50			−12

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 550 + (−1250) × 20 = 545.2

Answer: ₹545.20

Q3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.
Daily pocket allowance (₹) 11 − 13 13 − 15 15 − 17 17 − 19 19 − 21 21 − 23 23 − 25
Number of children 7 6 9 13 f 5 4

Daily pocket allowance (₹)	11 − 13	13 − 15	15 − 17	17 − 19	19 − 21	21 − 23	23 − 25
Number of children	7	6	9	13	f	5	4

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Class marks: 12, 14, 16, 18, 20, 22, 24. Σfᵢ = 44 + f.

Σfᵢxᵢ = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f.

752 + 20f44 + f = 18 ⇒ 752 + 20f = 792 + 18f ⇒ 2f = 40.

Answer: f = 20

Q4Thirty 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.

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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

Step-deviation method with a = 75.5, h = 3, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	30			4

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 75.5 + (430) × 3 = 75.9

Answer: 75.9 beats per minute

Q5In 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. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

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Number of mangoes	50 − 52	53 − 55	56 − 58	59 − 61	62 − 64
Number of boxes	15	110	135	115	25

Step-deviation method with a = 57, h = 3, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
50 − 52	15	51	−2	−30
53 − 55	110	54	−1	−110
56 − 58	135	57	0	0
59 − 61	115	60	1	115
62 − 64	25	63	2	50
Total	400			25

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 57 + (25400) × 3 = 57.19

The step-deviation method is chosen because the frequencies are large and the class marks are equally spaced (h = 3).

Answer: ≈ 57.19 mangoes (step-deviation method)

Q6The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

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Daily expenditure (₹)	100 − 150	150 − 200	200 − 250	250 − 300	300 − 350
Number of households	4	5	12	2	2

Step-deviation method with a = 225, h = 50, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	25			−7

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 225 + (−725) × 50 = 211

Answer: ₹211

Q7To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below. Find the mean concentration of SO₂ in the air.

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Concentration of SO₂ (ppm)	0 − 0.04	0.04 − 0.08	0.08 − 0.12	0.12 − 0.16	0.16 − 0.2	0.2 − 0.24
Frequency	4	9	9	2	4	2

Step-deviation method with a = 0.14, h = 0.04, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
0 − 0.04	4	0.02	−3	−12
0.04 − 0.08	9	0.06	−2	−18
0.08 − 0.12	9	0.1	−1	−9
0.12 − 0.16	2	0.14	0	0
0.16 − 0.2	4	0.18	1	4
0.2 − 0.24	2	0.22	2	4
Total	30			−31

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 0.14 + (−3130) × 0.04 = 0.10

Answer: ≈ 0.099 ppm

Q8A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

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Number of days	0 − 6	6 − 10	10 − 14	14 − 20	20 − 28	28 − 38	38 − 40
Number of students	11	10	7	4	4	3	1

Class	fᵢ	xᵢ	fᵢxᵢ
0 − 6	11	3	33
6 − 10	10	8	80
10 − 14	7	12	84
14 − 20	4	17	68
20 − 28	4	24	96
28 − 38	3	33	99
38 − 40	1	39	39
Total	40		499

x̄ = ΣfᵢxᵢΣfᵢ = 49940 = 12.475

Answer: 12.475 ≈ 12.48 days

Q9The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

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Literacy rate (%)	45 − 55	55 − 65	65 − 75	75 − 85	85 − 95
Number of cities	3	10	11	8	3

Step-deviation method with a = 70, h = 10, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
45 − 55	3	50	−2	−6
55 − 65	10	60	−1	−10
65 − 75	11	70	0	0
75 − 85	8	80	1	8
85 − 95	3	90	2	6
Total	35			−2

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 70 + (−235) × 10 = 69.43

Answer: ≈ 69.43%

Ex 13.2Exercise 13.2Mode of grouped data · 6 questions

Q1The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

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Age (in years)	5 − 15	15 − 25	25 − 35	35 − 45	45 − 55	55 − 65
Number of patients	6	11	21	23	14	5

The highest frequency is 23, so the modal class is 35 − 45. l = 35, h = 10, f₁ = 23, f₀ = 21, f₂ = 14.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 35 + (23 − 2146 − 21 − 14) × 10 = 36.82

Step-deviation method with a = 30, h = 10, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
5 − 15	6	10	−2	−12
15 − 25	11	20	−1	−11
25 − 35	21	30	0	0
35 − 45	23	40	1	23
45 − 55	14	50	2	28
55 − 65	5	60	3	15
Total	80			43

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 30 + (4380) × 10 = 35.375

The most patients admitted are about 36.8 years old (mode), while the average age of a patient is 35.375 years (mean).

Answer: Mode ≈ 36.82 years; mean = 35.375 years

Q2The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components.

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Lifetimes (in hours)	0 − 20	20 − 40	40 − 60	60 − 80	80 − 100	100 − 120
Frequency	10	35	52	61	38	29

The highest frequency is 61, so the modal class is 60 − 80. l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 60 + (61 − 52122 − 52 − 38) × 20 = 65.625

Answer: 65.625 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.

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Expenditure (₹)	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

The highest frequency is 40, so the modal class is 1500 − 2000. l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 1500 + (40 − 2480 − 24 − 33) × 500 = 1847.83

Step-deviation method with a = 2750, h = 500, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	200			−35

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 2750 + (−35200) × 500 = 2662.5

Answer: Mode ≈ ₹1847.83; mean = ₹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.

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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

The highest frequency is 10, so the modal class is 30 − 35. l = 30, h = 5, f₁ = 10, f₀ = 9, f₂ = 3.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 30 + (10 − 920 − 9 − 3) × 5 = 30.625

Step-deviation method with a = 32.5, h = 5, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	35			−23

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 32.5 + (−2335) × 5 = 29.21

Most states have about 30.6 students per teacher (mode); on average there are about 29.2 students per teacher (mean).

Answer: Mode = 30.625 ≈ 30.6; mean ≈ 29.21

Q5The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

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Runs scored	3000 − 4000	4000 − 5000	5000 − 6000	6000 − 7000	7000 − 8000	8000 − 9000	9000 − 10000	10000 − 11000
Number of batsmen	4	18	9	7	6	3	1	1

The highest frequency is 18, so the modal class is 4000 − 5000. l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 4000 + (18 − 436 − 4 − 9) × 1000 = 4608.70

Answer: ≈ 4608.70 runs

Q6A 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 given below. Find the mode of the data.

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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

The highest frequency is 20, so the modal class is 40 − 50. l = 40, h = 10, f₁ = 20, f₀ = 12, f₂ = 11.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 40 + (20 − 1240 − 12 − 11) × 10 = 44.71

Answer: ≈ 44.71 cars

Ex 13.3Exercise 13.3Median of grouped data · 7 questions

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.

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Monthly consumption (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

Class	Frequency	Cumulative frequency
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

n = 68, n/2 = 34. The first cumulative frequency greater than 34 is 42, so the median class is 125 − 145: l = 125, cf = 22, f = 20, h = 20.

Median = l + (n/2 − cff) × h = 125 + (34 − 2220) × 20 = 137

Step-deviation method with a = 135, h = 20, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	68			7

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 135 + (768) × 20 = 137.06

The highest frequency is 20, so the modal class is 125 − 145. l = 125, h = 20, f₁ = 20, f₀ = 13, f₂ = 14.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 125 + (20 − 1340 − 13 − 14) × 20 = 135.77

The three measures are close to one another (about 136 to 137 units), so the data is nearly symmetric.

Answer: Median 137 units; mean ≈ 137.06 units; mode ≈ 135.77 units

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 Total
Frequency 5 x 20 15 y 5 60

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Total: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15.

n/2 = 30. The median 28.5 lies in 20 − 30, so l = 20, f = 20, cf = 5 + x, h = 10.

28.5 = 20 + 30 − (5 + x)20 × 10 ⇒ 8.5 = 25 − x2 ⇒ x = 8.

Then y = 15 − 8 = 7.

Answer: 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 year.
Age (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

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These are cumulative frequencies. Taking the first class as 15 − 20 (2 holders, all aged 18 or 19) and subtracting consecutive entries:

Class	Frequency	Cumulative frequency
15 − 20	2	2
20 − 25	4	6
25 − 30	18	24
30 − 35	21	45
35 − 40	33	78
40 − 45	11	89
45 − 50	3	92
50 − 55	6	98
55 − 60	2	100

n = 100, n/2 = 50. The first cumulative frequency greater than 50 is 78, so the median class is 35 − 40: l = 35, cf = 45, f = 33, h = 5.

Median = l + (n/2 − cff) × h = 35 + (50 − 4533) × 5 = 35.76

Answer: ≈ 35.76 years

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. Find the median length of the leaves. (The classes 118 − 126, 127 − 135, … are first made continuous: 117.5 − 126.5, 126.5 − 135.5, …)

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Length (mm), continuous classes	117.5 − 126.5	126.5 − 135.5	135.5 − 144.5	144.5 − 153.5	153.5 − 162.5	162.5 − 171.5	171.5 − 180.5
Number of leaves	3	5	9	12	5	4	2

Class	Frequency	Cumulative frequency
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. The first cumulative frequency greater than 20 is 29, so the median class is 144.5 − 153.5: l = 144.5, cf = 17, f = 12, h = 9.

Median = l + (n/2 − cff) × h = 144.5 + (20 − 1712) × 9 = 146.75

Answer: 146.75 mm

Q5The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

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Life time (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

Class	Frequency	Cumulative frequency
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. The first cumulative frequency greater than 200 is 216, so the median class is 3000 − 3500: l = 3000, cf = 130, f = 86, h = 500.

Median = l + (n/2 − cff) × h = 3000 + (200 − 13086) × 500 = 3406.98

Answer: ≈ 3406.98 hours

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. 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.

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Number of letters	1 − 4	4 − 7	7 − 10	10 − 13	13 − 16	16 − 19
Number of surnames	6	30	40	16	4	4

Class	Frequency	Cumulative frequency
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. The first cumulative frequency greater than 50 is 76, so the median class is 7 − 10: l = 7, cf = 36, f = 40, h = 3.

Median = l + (n/2 − cff) × h = 7 + (50 − 3640) × 3 = 8.05

Step-deviation method with a = 8.5, h = 3, uᵢ = (xᵢ − a)/h.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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
Total	100			−6

x̄ = a + (ΣfᵢuᵢΣfᵢ) × h = 8.5 + (−6100) × 3 = 8.32

The highest frequency is 40, so the modal class is 7 − 10. l = 7, h = 3, f₁ = 40, f₀ = 30, f₂ = 16.

Mode = l + (f₁ − f₀2f₁ − f₀ − f₂) × h = 7 + (40 − 3080 − 30 − 16) × 3 = 7.88

Answer: Median ≈ 8.05; mean = 8.32; mode ≈ 7.88

Q7The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

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Weight (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

Class	Frequency	Cumulative frequency
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. The first cumulative frequency greater than 15 is 19, so the median class is 55 − 60: l = 55, cf = 13, f = 6, h = 5.

Median = l + (n/2 − cff) × h = 55 + (15 − 136) × 5 = 56.67

Answer: ≈ 56.67 kg

Class mark

Three methods

Example

Finding a missing frequency

Modal class

Formula

When to use which

Cumulative frequency

Formula

Continuous classes

Empirical relation

Formulas

Common slips

Q1Direct method

Q2Step deviation

Q3Mean of heights

Q4Assumed mean

Q5Missing frequency

Q6Mode

Q7Mode of heights

Q8Mode of earnings

Q9Mode of small data

Q10Median

Q11Median of heights

Q12All three

Q13Median of earnings

Q14Median of marks

Q15Empirical relation

Q16Less-than table

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. Which method did you use for finding the mean, and why?

Q2Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Q3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.Daily pocket allowance (₹)11 − 1313 − 1515 − 1717 − 1919 − 2121 − 2323 − 25Number of children76913f54

Q4Thirty 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.

Q6The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Q7To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below. Find the mean concentration of SO₂ in the air.

Q8A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Q9The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Q1The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Q2The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components.

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.

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.

Q5The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

Q6A 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 given below. Find the mode of the 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.

Q2If the median of the distribution given below is 28.5, find the values of x and y.Class interval0 − 1010 − 2020 − 3030 − 4040 − 5050 − 60TotalFrequency5x2015y560

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. Find the median length of the leaves. (The classes 118 − 126, 127 − 135, … are first made continuous: 117.5 − 126.5, 126.5 − 135.5, …)

Q5The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

Q7The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Q3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.
Daily pocket allowance (₹) 11 − 13 13 − 15 15 − 17 17 − 19 19 − 21 21 − 23 23 − 25
Number of children 7 6 9 13 f 5 4

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 Total
Frequency 5 x 20 15 y 5 60