📊 Class 10 NCERT Mathematics

Chapter 13: Statistics - Complete Study Guide

📝 Important Notes & Formulas — Chapter 13

📐 Mean of Grouped Data

Three Methods to Calculate:

Direct Method:
x̄ = Σfᵢxᵢ / Σfᵢ where xᵢ = class mark (mid-point), fᵢ = frequency
Assumed Mean Method:
x̄ = a + Σfᵢdᵢ / Σfᵢ where dᵢ = xᵢ − a, a = assumed mean
Step Deviation Method:
x̄ = a + (Σfᵢuᵢ/Σfᵢ) × h where uᵢ = (xᵢ − a)/h, h = class width

💡 When to use: Step Deviation is best when class widths are equal and data values are large.

📈 Mode of Grouped Data

Modal Class: The class interval with the highest frequency.
Formula:
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
Terms:
- l = lower limit of modal class
- h = class size
- f₁ = frequency of modal class
- f₀ = frequency of class preceding modal class
- f₂ = frequency of class succeeding modal class

📉 Median of Grouped Data

Preparation: Classes must be continuous (exclusive form) before applying formula.
Steps:
1. Find n/2 (where n = Σfᵢ)
2. Identify median class — the class whose cumulative frequency is ≥ n/2
Formula:
Median = l + [(n/2 − cf)/f] × h
Terms:
- l = lower limit of median class
- cf = cumulative frequency of class preceding median class
- f = frequency of median class
- h = class size

📊 Cumulative Frequency & Ogives

Less than type: Add frequencies from top to bottom. Plot upper class limits vs cumulative frequency.
More than type: Add frequencies from bottom to top. Plot lower class limits vs cumulative frequency.
Ogive (Cumulative Frequency Curve): A smooth curve drawn through these points.
Finding Median from Ogive: The x-coordinate of the intersection point of less than and more than ogives gives the median.

🔗 Empirical Relationship

For a moderately skewed distribution:

Mode = 3 × Median − 2 × Mean

Or equivalently:

3 Median = Mode + 2 Mean

Skewness Check:

If Mean > Median > Mode → Positively skewed (right tail)
If Mean < Median < Mode → Negatively skewed (left tail)
If Mean ≈ Median ≈ Mode → Symmetrical distribution

⚠️ Critical Points & Exam Tips

Inclusive to Exclusive: For inclusive classes (e.g., 10–19, 20–29), convert to continuous by subtracting 0.5 from lower limits and adding 0.5 to upper limits.
Cumulative Frequency Tables: Always convert to regular frequencies first before calculating mean or mode.
Unequal Class Widths: Adjusted frequency = (minimum class width / class width) × given frequency. Use adjusted frequencies for histograms.
Check n: Always verify that Σf = total number of observations given in the question.
Decimal Accuracy: Round final answers to 2 decimal places unless specified otherwise.
Missing Frequency Problems: Use the given mean/median value to form an equation and solve for the unknown frequency.

📐 Exercise 13.1 — Mean of Grouped Data

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Which method did you use for finding the mean, and why?

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

A Mean = 7.5 (Assumed Mean Method)

B Mean = 8.1 (Direct Method)

C Mean = 8.5 (Step-Deviation Method)

D Mean = 7.8 (Direct Method)

✅ Solution:

Step 1: Mid-points: 1, 3, 5, 7, 9, 11, 13

Step 2: Calculate fx: 1×1=1, 2×3=6, 1×5=5, 5×7=35, 6×9=54, 2×11=22, 3×13=39

Step 3: Σfx = 162, Σf = 20

Step 4: Mean = 162/20 = 8.1 plants per house

Method used: Direct Method (simple calculations, small data)

Consider 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.

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

A ₹ 540.0

B ₹ 550.0

C ₹ 545.2

D ₹ 538.5

✅ Solution:

Step 1: Mid-points: 510, 530, 550, 570, 590

Step 2: fx: 12×510=6120, 14×530=7420, 8×550=4400, 6×570=3420, 10×590=5900

Step 3: Σfx = 27260, Σf = 50

Step 4: Mean = 27260/50 = ₹ 545.2

The 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 (in ₹)	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Number of children	7	6	9	13	f	5	4

A f = 18

B f = 22

C f = 20

D f = 24

✅ Solution:

Step 1: Mid-points: 12, 14, 16, 18, 20, 22, 24

Step 2: Σf = 44 + f, Σfx = 752 + 20f

Step 3: 18 = (752 + 20f)/(44 + f)

Step 4: 792 + 18f = 752 + 20f → 40 = 2f → f = 20

Thirty 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

A 74.5 beats/min

B 76.2 beats/min

C 72.8 beats/min

D 75.9 beats/min

✅ Solution:

Step 1: Mid-points: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5

Step 2: fx: 133, 278, 217.5, 604, 549.5, 326, 169

Step 3: Σfx = 2277, Σf = 30

Step 4: Mean = 2277/30 = 75.9 beats/min

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

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

A 56.5 (Direct Method)

B 57.19 (Direct Method)

C 57.19 (Assumed Mean)

D 58.0 (Step-Deviation)

✅ Solution:

Step 1: Mid-points: 51, 54, 57, 60, 63

Step 2: fx: 765, 5940, 7695, 6900, 1575

Step 3: Σfx = 22875, Σf = 400

Step 4: Mean = 22875/400 = 57.19 mangoes

Method: Direct Method (easy multiplication)

The 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.

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

A ₹ 205

B ₹ 211

C ₹ 220

D ₹ 198

✅ Solution:

Step 1: Mid-points: 125, 175, 225, 275, 325

Step 2: fx: 500, 875, 2700, 550, 650

Step 3: Σfx = 5275, Σf = 25

Step 4: Mean = 5275/25 = ₹ 211

To 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.

Concentration of SO₂ (in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

A 0.12 ppm

B 0.099 ppm

C 0.085 ppm

D 0.105 ppm

✅ Solution:

Step 1: Mid-points: 0.02, 0.06, 0.10, 0.14, 0.18, 0.22

Step 2: fx: 0.08, 0.54, 0.90, 0.28, 0.72, 0.44

Step 3: Σfx = 2.96, Σf = 30

Step 4: Mean = 2.96/30 = 0.099 ppm

A 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.

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

A 11.2 days

B 13.5 days

C 12.48 days

D 10.8 days

✅ Solution:

Step 1: Mid-points: 3, 8, 12, 17, 24, 33, 39

Step 2: fx: 33, 80, 84, 68, 96, 99, 39

Step 3: Σfx = 499, Σf = 40

Step 4: Mean = 499/40 = 12.48 days

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

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

A 68.5%

B 72.1%

C 69.43%

D 65.8%

✅ Solution:

Step 1: Mid-points: 50, 60, 70, 80, 90

Step 2: fx: 150, 600, 770, 640, 270

Step 3: Σfx = 2430, Σf = 35

Step 4: Mean = 2430/35 = 69.43%

🧮 Mean Calculator (Practice)

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📈 Exercise 13.2 — Mode and Mean of Grouped Data

The 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.

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

A Mode = 35.5, Mean = 34.2

B Mode = 36.82, Mean = 35.38

C Mode = 38.2, Mean = 36.5

D Mode = 34.5, Mean = 33.8

✅ Solution:

Mode: Modal class = 35–45 (f=23)

Mode = 35 + [(23−21)/(46−21−14)] × 10 = 35 + 1.82 = 36.82 years

Mean: Σfx = 2830, Σf = 80

Mean = 2830/80 = 35.38 years

Interpretation: Mode (36.82) > Mean (35.38), data slightly skewed towards younger ages

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

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

A 60.5 hours

B 70.2 hours

C 65.625 hours

D 62.0 hours

✅ Solution:

Step 1: Modal class = 60–80 (f=61)

Step 2: l=60, h=20, f₁=61, f₀=52, f₂=38

Step 3: Mode = 60 + [(61−52)/(122−52−38)] × 20

Step 4: Mode = 60 + (9/32) × 20 = 65.625 hours

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Expenditure (in ₹)	Number of families
1000 – 1500	24
1500 – 2000	40
2000 – 2500	33
2500 – 3000	28
3000 – 3500	30
3500 – 4000	22
4000 – 4500	16
4500 – 5000	7

A Mode = ₹1800, Mean = ₹2500

B Mode = ₹1750, Mean = ₹2600

C Mode = ₹1847.83, Mean = ₹2662.50

D Mode = ₹1900, Mean = ₹2700

✅ Solution:

Mode: Modal class = 1500–2000 (f=40)

Mode = 1500 + [(40−24)/(80−24−33)] × 500 = 1500 + 347.83 = ₹1847.83

Mean: Σfx = 532500, Σf = 200

Mean = 532500/200 = ₹2662.50

The 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	Number of states/U.T.
15 – 20	3
20 – 25	8
25 – 30	9
30 – 35	10
35 – 40	3
40 – 45	0
45 – 50	0
50 – 55	2

A Mode = 32.5, Mean = 30.5

B Mode = 28.2, Mean = 27.5

C Mode = 30.625, Mean = 29.21

D Mode = 31.0, Mean = 28.8

✅ Solution:

Mode: Modal class = 30–35 (f=10)

Mode = 30 + [(10−9)/(20−9−3)] × 5 = 30 + 0.625 = 30.625

Mean: Σfx = 1022.5, Σf = 35

Mean = 1022.5/35 = 29.21

Interpretation: Most common ratio is ~30.6, average is ~29.2

The 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.

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

A 4200 runs

B 4800 runs

C 4608.7 runs

D 4500 runs

✅ Solution:

Step 1: Modal class = 4000–5000 (f=18)

Step 2: l=4000, h=1000, f₁=18, f₀=4, f₂=9

Step 3: Mode = 4000 + [(18−4)/(36−4−9)] × 1000

Step 4: Mode = 4000 + (14/23) × 1000 = 4608.7 runs

A 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.

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

A 42.5 cars

B 46.2 cars

C 43.8 cars

D 44.71 cars

✅ Solution:

Step 1: Modal class = 40–50 (f=20)

Step 2: l=40, h=10, f₁=20, f₀=12, f₂=11

Step 3: Mode = 40 + [(20−12)/(40−12−11)] × 10

Step 4: Mode = 40 + (8/17) × 10 = 44.71 cars

🧮 Mode Calculator (Practice)

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📉 Exercise 13.3 — Median, Mean and Mode of Grouped Data

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	Number of consumers
65 – 85	4
85 – 105	5
105 – 125	13
125 – 145	20
145 – 165	14
165 – 185	8
185 – 205	4

A Median=135, Mean=138, Mode=140

B Median=137, Mean=137.06, Mode=135.77

C Median=140, Mean=136, Mode=138

D Median=130, Mean=135, Mode=132

✅ Solution:

Median: n=68, n/2=34, Median class=125–145

Median = 125 + [(34−22)/20] × 20 = 137 units

Mean: Σfx=9320, Σf=68, Mean=9320/68=137.06

Mode: Modal class=125–145, Mode=125+[(20−13)/(40−13−14)]×20=135.77

Comparison: Mean≈Median≈Mode, distribution is approximately symmetrical

If the median of the distribution given below is 28.5, find the values of x and y.

Class interval	Frequency
0 – 10	5
10 – 20	x
20 – 30	20
30 – 40	15
40 – 50	y
50 – 60	5
Total	60

A x = 6, y = 9

B x = 10, y = 5

C x = 8, y = 7

D x = 7, y = 8

✅ Solution:

Step 1: 5 + x + 20 + 15 + y + 5 = 60 → x + y = 15

Step 2: Median=28.5 lies in class 20–30

Step 3: 28.5 = 20 + [(30−(5+x))/20] × 10

Step 4: 8.5 = (25−x)/2 → 17 = 25−x → x = 8

Step 5: y = 15 − 8 = 7

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)	Number of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

A 34.2 years

B 36.5 years

C 35.76 years

D 33.8 years

✅ Solution:

Step 1: Convert to regular frequencies: 18–20:2, 20–25:4, 25–30:18, 30–35:21, 35–40:33, 40–45:11, 45–50:3, 50–55:6, 55–60:2

Step 2: n=100, n/2=50, Median class=35–40 (cf=45)

Step 3: Median = 35 + [(50−45)/33] × 5

Step 4: Median = 35 + 25/33 = 35.76 years

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves. (Hint: Convert to continuous classes: 117.5–126.5, 126.5–135.5, ..., 171.5–180.5)

Length (in mm)	Number of leaves
118 – 126	3
127 – 135	5
136 – 144	9
145 – 153	12
154 – 162	5
163 – 171	4
172 – 180	2

A 144.5 mm

B 148.0 mm

C 146.75 mm

D 145.2 mm

✅ Solution:

Step 1: Convert to continuous: 117.5–126.5:3, 126.5–135.5:5, 135.5–144.5:9, 144.5–153.5:12, 153.5–162.5:5, 162.5–171.5:4, 171.5–180.5:2

Step 2: n=40, n/2=20, Median class=144.5–153.5 (cf=17)

Step 3: Median = 144.5 + [(20−17)/12] × 9

Step 4: Median = 144.5 + 2.25 = 146.75 mm

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

Life time (in hours)	Number of lamps
1500 – 2000	14
2000 – 2500	56
2500 – 3000	60
3000 – 3500	86
3500 – 4000	74
4000 – 4500	62
4500 – 5000	48

A 3200 hours

B 3500 hours

C 3406.98 hours

D 3300 hours

✅ Solution:

Step 1: Build cf: 14, 70, 130, 216, 290, 352, 400

Step 2: n=400, n/2=200, Median class=3000–3500 (cf=130)

Step 3: Median = 3000 + [(200−130)/86] × 500

Step 4: Median = 3000 + 406.98 = 3406.98 hours

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows. 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.

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

A Median=8.2, Mean=8.5, Mode=8.0

B Median=8.8, Mean=9.0, Mode=8.5

C Median=8.55, Mean=8.79, Mode=8.38

D Median=8.0, Mean=8.6, Mode=8.2

✅ Solution:

Convert to continuous: 0.5–4.5:6, 4.5–7.5:30, 7.5–10.5:40, 10.5–13.5:16, 13.5–16.5:4, 16.5–19.5:4

Median: n=100, n/2=50, class=7.5–10.5, Median=7.5+[(50−36)/40]×3=8.55

Mean: Σfx=879, Σf=100, Mean=8.79

Mode: Modal class=7.5–10.5, Mode=7.5+[(40−30)/(80−30−16)]×3=8.38

The 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

A 55.0 kg

B 58.5 kg

C 56.67 kg

D 54.2 kg

✅ Solution:

Step 1: Build cf: 2, 5, 13, 19, 25, 28, 30

Step 2: n=30, n/2=15, Median class=55–60 (cf=13)

Step 3: Median = 55 + [(15−13)/6] × 5

Step 4: Median = 55 + 10/6 = 56.67 kg

🧮 Median Calculator (Practice)

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