Chapter 13 of 16

Updated 6 August 2026

NCERT Solutions

Statistics — NCERT Solutions

CBSE · Class 10 · Mathematics

NCERT Solutions for Statistics, CBSE Class 10 Mathematics: 22 textbook questions solved step by step. Part of the CBSE Class 10 Mathematics syllabus.

160 questions56 flashcards11 formulas & key relations5 concepts

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22 Questions Solved · 3 Sections

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Exercise 13.1

1A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

| Number of plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |
|---|---|---|---|---|---|---|---|
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |

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

Given: Frequency distribution of number of plants in 20 houses.

Method Used: Direct Method (since the values of $x_i$ are small and easy to compute).

Formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

Step 1: Find the midpoint $x_i$ of each class.

Class	$f_i$	$x_i$ (midpoint)	$f_i x_i$
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

Step 2: Calculate the mean.
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1$

Answer: The mean number of plants per house is 8.1.

Reason for choosing Direct Method: The class midpoints $x_i$ are small integers, making direct multiplication simple without needing an assumed mean.

2Consider the following distribution of daily wages of 50 workers of a factory.

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

Find the mean daily wages of the workers of the factory by using an appropriate method.Show solution

Given: Frequency distribution of daily wages of 50 workers.

Method Used: Assumed Mean Method (since the values of $x_i$ are large).

Let assumed mean $a = 550$ , class width $h = 20$ .

Formula: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad \text{where } d_i = x_i - a$

Step 1: Prepare the table.

Class	$f_i$	$x_i$	$d_i = x_i - 550$	$f_i d_i$
500 – 520	12	510	–40	–480
520 – 540	14	530	–20	–280
540 – 560	8	550	0	0
560 – 580	6	570	20	120
580 – 600	10	590	40	400
Total	50			–240

Step 2: Calculate the mean.
$\bar{x} = 550 + \frac{-240}{50} = 550 - 4.8 = 545.20$

Answer: The mean daily wages of the workers is ₹ 545.20.

3The 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 ₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
|---|---|---|---|---|---|---|---|
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |Show solution

Given: Mean $\bar{x} = 18$ , one frequency is missing.

Method: Direct Method.

Formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

Step 1: Prepare the table.

Class	$f_i$	$x_i$ (midpoint)	$f_i x_i$
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
Total	$44 + f$		$752 + 20f$

Step 2: Apply the mean formula.
$18 = \frac{752 + 20f}{44 + f}$

$18(44 + f) = 752 + 20f$

$792 + 18f = 752 + 20f$

$792 - 752 = 20f - 18f$

$40 = 2f$

$f = 20$

Answer: The missing frequency $f$ = 20.

4Thirty 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 |Show solution

Given: Frequency distribution of heartbeats per minute for 30 women.

Method Used: Assumed Mean Method.

Let assumed mean $a = 75.5$ , class width $h = 3$ .

Formula: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad d_i = x_i - a$

Step 1: Prepare the table.

Class	$f_i$	$x_i$	$d_i = x_i - 75.5$	$f_i d_i$
65 – 68	2	66.5	–9	–18
68 – 71	4	69.5	–6	–24
71 – 74	3	72.5	–3	–9
74 – 77	8	75.5	0	0
77 – 80	7	78.5	3	21
80 – 83	4	81.5	6	24
83 – 86	2	84.5	9	18
Total	30			12

Step 2: Calculate the mean.
$\bar{x} = 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$

Answer: The mean heartbeats per minute for these women is 75.9 beats per minute.

5In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

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

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

Given: The classes are not continuous (gaps exist: 50–52, 53–55, …). We first note that these are inclusive classes. The midpoints are taken directly.

Method Used: Step Deviation Method (since frequencies are large and values are close together).

Let assumed mean $a = 57$ , class width $h = 3$ .

Formula: $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h, \quad u_i = \frac{x_i - a}{h}$

Step 1: Find midpoints and prepare the table.

Class	$f_i$	$x_i$	$u_i = \frac{x_i-57}{3}$	$f_i u_i$
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

Step 2: Calculate the mean.
$\bar{x} = 57 + \frac{25}{400} \times 3 = 57 + \frac{75}{400} = 57 + 0.1875 = 57.19 \text{ (approx.)}$

Answer: The mean number of mangoes kept in a packing box is 57.19 (approximately).

Reason: Step deviation method was chosen because the frequencies are large and the class widths are equal, making calculations simpler.

6The table below shows the daily expenditure on food of 25 households in a locality.

| Daily expenditure (in ₹) | 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.Show solution

Given: Frequency distribution of daily food expenditure of 25 households.

Method Used: Step Deviation Method.

Let assumed mean $a = 225$ , class width $h = 50$ .

Formula: $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h, \quad u_i = \frac{x_i - a}{h}$

Step 1: Prepare the table.

Class	$f_i$	$x_i$	$u_i = \frac{x_i - 225}{50}$	$f_i u_i$
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

Step 2: Calculate the mean.
$\bar{x} = 225 + \frac{-7}{25} \times 50 = 225 - 14 = 211$

Answer: The mean daily expenditure on food is ₹ 211.

7To 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:

| 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 |

Find the mean concentration of SO₂ in the air.Show solution

Given: Frequency distribution of SO₂ concentration for 30 localities.

Method Used: Direct Method.

Formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

Step 1: Find midpoints and prepare the table.

Class	$f_i$	$x_i$ (midpoint)	$f_i x_i$
0.00 – 0.04	4	0.02	0.08
0.04 – 0.08	9	0.06	0.54
0.08 – 0.12	9	0.10	0.90
0.12 – 0.16	2	0.14	0.28
0.16 – 0.20	4	0.18	0.72
0.20 – 0.24	2	0.22	0.44
Total	30		2.96

Step 2: Calculate the mean.
$\bar{x} = \frac{2.96}{30} = 0.0987 \text{ ppm (approx.)}$

Answer: The mean concentration of SO₂ in the air is 0.099 ppm (approximately).

8A 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 |Show solution

Given: Frequency distribution of absentee days for 40 students.

Method Used: Direct Method.

Formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

Step 1: Find midpoints and prepare the table.

Class	$f_i$	$x_i$ (midpoint)	$f_i x_i$
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

Step 2: Calculate the mean.
$\bar{x} = \frac{499}{40} = 12.475$

Answer: The mean number of days a student was absent is 12.48 days (approximately).

9The 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 |Show solution

Given: Frequency distribution of literacy rates of 35 cities.

Method Used: Step Deviation Method.

Let assumed mean $a = 70$ , class width $h = 10$ .

Formula: $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h, \quad u_i = \frac{x_i - a}{h}$

Step 1: Prepare the table.

Class	$f_i$	$x_i$	$u_i = \frac{x_i - 70}{10}$	$f_i u_i$
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

Step 2: Calculate the mean.
$\bar{x} = 70 + \frac{-2}{35} \times 10 = 70 - \frac{20}{35} = 70 - \frac{4}{7} = 70 - 0.571 = 69.43 \text{ (approx.)}$

Answer: The mean literacy rate is 69.43% (approximately).

Exercise 13.2

1The 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.Show solution

Given: Frequency distribution of ages of patients.

FINDING MODE:

Step 1: Identify the modal class (class with highest frequency).

Highest frequency = 23, which belongs to class 35 – 45.

So, Modal class = 35 – 45.

$l = 35$ , $f_1 = 23$ , $f_0 = 21$ , $f_2 = 14$ , $h = 10$

Formula: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

$\text{Mode} = 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10 = 35 + \frac{2}{46 - 35} \times 10 = 35 + \frac{2}{11} \times 10$

$\text{Mode} = 35 + \frac{20}{11} = 35 + 1.81 = 36.8 \text{ years (approx.)}$

FINDING MEAN:

Let assumed mean $a = 30$ , $h = 10$ .

Class	$f_i$	$x_i$	$u_i = \frac{x_i-30}{10}$	$f_i u_i$
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

$\bar{x} = 30 + \frac{43}{80} \times 10 = 30 + 5.375 = 35.375 \approx 35.38 \text{ years}$

Interpretation:

Mean age ≈ 35.38 years — this is the average age of all patients.
Mode ≈ 36.8 years — this is the age group most commonly admitted.

Both values are close, indicating that the maximum number of patients admitted are in the age group around 35–45 years.

2The 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.Show solution

Given: Frequency distribution of lifetimes of 225 electrical components.

Step 1: Identify the modal class.

Highest frequency = 61, which belongs to class 60 – 80.

Modal class = 60 – 80.

$l = 60$ , $f_1 = 61$ , $f_0 = 52$ , $f_2 = 38$ , $h = 20$

Formula: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

$\text{Mode} = 60 + \frac{61 - 52}{2(61) - 52 - 38} \times 20$

$= 60 + \frac{9}{122 - 90} \times 20 = 60 + \frac{9}{32} \times 20$

$= 60 + \frac{180}{32} = 60 + 5.625 = 65.625 \text{ hours}$

Answer: The modal lifetime of the components is 65.625 hours.

3The 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 |

Given: Frequency distribution of monthly expenditure of 200 families.

FINDING MODE:

Highest frequency = 40, Modal class = 1500 – 2000.

$l = 1500$ , $f_1 = 40$ , $f_0 = 24$ , $f_2 = 33$ , $h = 500$

$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

$= 1500 + \frac{40 - 24}{2(40) - 24 - 33} \times 500 = 1500 + \frac{16}{80 - 57} \times 500$

$= 1500 + \frac{16}{23} \times 500 = 1500 + \frac{8000}{23} = 1500 + 347.83 = 1847.83$

Modal monthly expenditure ≈ ₹ 1847.83

FINDING MEAN:

Let assumed mean $a = 3250$ , $h = 500$ .

Class	$f_i$	$x_i$	$u_i = \frac{x_i - 3250}{500}$	$f_i u_i$
1000 – 1500	24	1250	–4	–96
1500 – 2000	40	1750	–3	–120
2000 – 2500	33	2250	–2	–66
2500 – 3000	28	2750	–1	–28
3000 – 3500	30	3250	0	0
3500 – 4000	22	3750	1	22
4000 – 4500	16	4250	2	32
4500 – 5000	7	4750	3	21
Total	200			–235

$\bar{x} = 3250 + \frac{-235}{200} \times 500 = 3250 - \frac{235 \times 500}{200} = 3250 - 587.5 = 2662.5$

Answer: Modal monthly expenditure = ₹ 1847.83 and Mean monthly expenditure = ₹ 2662.50.

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4The 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 |

Given: Frequency distribution of teacher-student ratio.

FINDING MODE:

Highest frequency = 10, Modal class = 30 – 35.

$l = 30$ , $f_1 = 10$ , $f_0 = 9$ , $f_2 = 3$ , $h = 5$

$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{10 - 9}{2(10) - 9 - 3} \times 5$

$= 30 + \frac{1}{20 - 12} \times 5 = 30 + \frac{5}{8} = 30 + 0.625 = 30.6 \text{ (approx.)}$

FINDING MEAN:

Let assumed mean $a = 32.5$ , $h = 5$ .

Class	$f_i$	$x_i$	$u_i = \frac{x_i - 32.5}{5}$	$f_i u_i$
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

$\bar{x} = 32.5 + \frac{-23}{35} \times 5 = 32.5 - \frac{115}{35} = 32.5 - 3.286 = 29.2 \text{ (approx.)}$

Interpretation:

Mode ≈ 30.6 students per teacher — the most common teacher-student ratio among states/UTs.
Mean ≈ 29.2 students per teacher — the average teacher-student ratio across all states/UTs.

The mode is slightly higher than the mean, indicating that more states have a ratio around 30–35 students per teacher.

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5The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

| 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 |

Find the mode of the data.

Given: Frequency distribution of runs scored by top batsmen.

Step 1: Identify the modal class.

Highest frequency = 18, Modal class = 4000 – 5000.

$l = 4000$ , $f_1 = 18$ , $f_0 = 4$ , $f_2 = 9$ , $h = 1000$

Formula: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

$= 4000 + \frac{18 - 4}{2(18) - 4 - 9} \times 1000 = 4000 + \frac{14}{36 - 13} \times 1000$

$= 4000 + \frac{14}{23} \times 1000 = 4000 + 608.7 = 4608.7 \text{ runs (approx.)}$

Answer: The mode of the data is 4608.7 runs (approximately).

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6A 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 |

Given: Frequency distribution of number of cars passing through a spot.

Step 1: Identify the modal class.

Highest frequency = 20, Modal class = 40 – 50.

$l = 40$ , $f_1 = 20$ , $f_0 = 12$ , $f_2 = 11$ , $h = 10$

Formula: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

$= 40 + \frac{20 - 12}{2(20) - 12 - 11} \times 10 = 40 + \frac{8}{40 - 23} \times 10$

$= 40 + \frac{8}{17} \times 10 = 40 + \frac{80}{17} = 40 + 4.7 = 44.7 \text{ (approx.)}$

Answer: The mode of the data is 44.7 cars (approximately).

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Exercise 13.3

1The 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 |

Given: Frequency distribution of monthly electricity consumption of 68 consumers.

FINDING MEDIAN:

$n = 68$ , so $\frac{n}{2} = 34$ .

Cumulative Frequency Table:

Class	$f_i$	Cumulative Frequency (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

The cumulative frequency just exceeding 34 is 42, so the median class = 125 – 145.

$l = 125$ , $cf = 22$ , $f = 20$ , $h = 20$

$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 125 + \frac{34 - 22}{20} \times 20 = 125 + \frac{12}{20} \times 20 = 125 + 12 = 137$

Median = 137 units

FINDING MODE:

Highest frequency = 20, Modal class = 125 – 145.

$l = 125$ , $f_1 = 20$ , $f_0 = 13$ , $f_2 = 14$ , $h = 20$

$\text{Mode} = 125 + \frac{20 - 13}{2(20) - 13 - 14} \times 20 = 125 + \frac{7}{40 - 27} \times 20 = 125 + \frac{7}{13} \times 20$

$= 125 + \frac{140}{13} = 125 + 10.77 = 135.77 \approx 135.8 \text{ units}$

FINDING MEAN:

Let assumed mean $a = 135$ , $h = 20$ .

Class	$f_i$	$x_i$	$u_i = \frac{x_i-135}{20}$	$f_i u_i$
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

$\bar{x} = 135 + \frac{7}{68} \times 20 = 135 + \frac{140}{68} = 135 + 2.06 = 137.06 \approx 137.05 \text{ units}$

Comparison:

Mean ≈ 137.05 units
Median = 137 units
Mode ≈ 135.8 units

All three values are very close to each other, indicating that the distribution is nearly symmetrical. The maximum number of consumers use electricity in the range 125–145 units.

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2If 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 |

Given: Median = 28.5, Total frequency = 60.

Step 1: Write the equation from total frequency.
$5 + x + 20 + 15 + y + 5 = 60$
$45 + x + y = 60$
$x + y = 15 \quad \cdots (1)$

Step 2: Since median = 28.5, the median class is 20 – 30.

$\frac{n}{2} = 30$

$l = 20$ , $cf = 5 + x$ , $f = 20$ , $h = 10$

Formula: $\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$

$28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10$

$28.5 - 20 = \frac{25 - x}{20} \times 10$

$8.5 = \frac{25 - x}{2}$

$17 = 25 - x$

$x = 8$

Step 3: Substitute in equation (1):
$8 + y = 15 \Rightarrow y = 7$

Answer: $x = \mathbf{8}$ and $y = \mathbf{7}$ .

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3A 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 |

Given: Cumulative frequency distribution of ages of 100 policy holders.

Step 1: Convert to class-wise frequency distribution.

Class (Age)	Frequency ( $f_i$ )	Cumulative Frequency (cf)
18 – 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

Step 2: $n = 100$ , $\frac{n}{2} = 50$ .

The cumulative frequency just exceeding 50 is 78, so the median class = 35 – 40.

$l = 35$ , $cf = 45$ , $f = 33$ , $h = 5$

Formula: $\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$

$= 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{5}{33} \times 5 = 35 + \frac{25}{33}$

$= 35 + 0.76 = 35.76 \text{ years (approx.)}$

Answer: The median age of the policy holders is 35.76 years (approximately).

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4The 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) | Number of leaves |
|---|---|
| 118 – 126 | 3 |
| 127 – 135 | 5 |
| 136 – 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ..., 171.5 – 180.5.)

Given: The data is in inclusive form. Convert to exclusive (continuous) form by subtracting 0.5 from lower limit and adding 0.5 to upper limit.

Step 1: Prepare the cumulative frequency table.

Class (continuous)	$f_i$	Cumulative Frequency (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

Step 2: $n = 40$ , $\frac{n}{2} = 20$ .

The cumulative frequency just exceeding 20 is 29, so the median class = 144.5 – 153.5.

$l = 144.5$ , $cf = 17$ , $f = 12$ , $h = 9$

Formula: $\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$

$= 144.5 + \frac{20 - 17}{12} \times 9 = 144.5 + \frac{3}{12} \times 9 = 144.5 + \frac{27}{12}$

$= 144.5 + 2.25 = 146.75 \text{ mm}$

Answer: The median length of the leaves is 146.75 mm.

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5The following table gives the distribution of the life time of 400 neon lamps:

| 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 |

Find the median life time of a lamp.

Given: Frequency distribution of lifetimes of 400 neon lamps.

Step 1: Prepare the cumulative frequency table.

Class	$f_i$	Cumulative Frequency (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

Step 2: $n = 400$ , $\frac{n}{2} = 200$ .

The cumulative frequency just exceeding 200 is 216, so the median class = 3000 – 3500.

$l = 3000$ , $cf = 130$ , $f = 86$ , $h = 500$

Formula: $\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$

$= 3000 + \frac{200 - 130}{86} \times 500 = 3000 + \frac{70}{86} \times 500$

$= 3000 + \frac{35000}{86} = 3000 + 406.98 = 3406.98 \text{ hours (approx.)}$

Answer: The median lifetime of a lamp is 3406.98 hours (approximately).

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6100 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.

Given: Frequency distribution of number of letters in 100 surnames.

FINDING MEDIAN:

Class	$f_i$	Cumulative Frequency (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

$n = 100$ , $\frac{n}{2} = 50$ .

The cumulative frequency just exceeding 50 is 76, so median class = 7 – 10.

$l = 7$ , $cf = 36$ , $f = 40$ , $h = 3$

$\text{Median} = 7 + \frac{50 - 36}{40} \times 3 = 7 + \frac{14}{40} \times 3 = 7 + \frac{42}{40} = 7 + 1.05 = 8.05$

Median = 8.05 letters

FINDING MEAN:

Let assumed mean $a = 8.5$ , $h = 3$ .

Class	$f_i$	$x_i$	$u_i = \frac{x_i - 8.5}{3}$	$f_i u_i$
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

$\bar{x} = 8.5 + \frac{-6}{100} \times 3 = 8.5 - 0.18 = 8.32$

Mean = 8.32 letters

FINDING MODE:

Highest frequency = 40, Modal class = 7 – 10.

$l = 7$ , $f_1 = 40$ , $f_0 = 30$ , $f_2 = 16$ , $h = 3$

$\text{Mode} = 7 + \frac{40 - 30}{2(40) - 30 - 16} \times 3 = 7 + \frac{10}{80 - 46} \times 3 = 7 + \frac{10}{34} \times 3$

$= 7 + \frac{30}{34} = 7 + 0.88 = 7.88$

Mode ≈ 7.88 letters

Answer: Median = 8.05, Mean = 8.32, Mode ≈ 7.88 letters.

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7The 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 |

Given: Frequency distribution of weights of 30 students.

Step 1: Prepare the cumulative frequency table.

Class	$f_i$	Cumulative Frequency (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

Step 2: $n = 30$ , $\frac{n}{2} = 15$ .

The cumulative frequency just exceeding 15 is 19, so the median class = 55 – 60.

$l = 55$ , $cf = 13$ , $f = 6$ , $h = 5$

Formula: $\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$

$= 55 + \frac{15 - 13}{6} \times 5 = 55 + \frac{2}{6} \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67 \text{ kg (approx.)}$

Answer: The median weight of the students is 56.67 kg (approximately).

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