NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.3

Q1. 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.

Difficulty Level: Medium
Known: Monthly consumption (in units) of 68 consumers (grouped data)
Unknown: Mean, Median, Mode and comparison

Given Table (Monthly consumption → Number of consumers):
65–85 → 4
85–105 → 5
105–125 → 13
125–145 → 20
145–165 → 14
165–185 → 8
185–205 → 4

Reasoning

Mean will be found using step-deviation method:

$\bar{X} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)$ $X ˉ = a + h (\sum f ^{i} \sum f ^{i} u ^{i})$
Mode will be found using modal class formula (highest frequency class):

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h$ $Mode = l + (2 f ^{1} - f ^{0} - f ^{2} f ^{1} - f ^{0}) h$
Median will be found using median formula, where median class is the class whose cumulative frequency is just greater than

$\frac{n}{2}$ $2 n$ :

$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right)h$ $Median = l + (f 2 n - c f) h$

Solution

1) Mean (Step-Deviation Method)

Class size (h) = 20
Assumed mean (a) = 135

We first calculate class marks

$x_i$ $x_{i}$ , then

$d_i = x_i - a$ $d_{i} = x_{i} - a$ , then

$u_i = \frac{d_i}{h}$ $u_{i} = h d ^{i}$ , and finally

$f_i u_i$ $f_{i} u_{i}$ .

Monthly consumption (units)	$f_i$ $f_{i}$	Class mark $x_i$ $x_{i}$	$d_i=x_i-a$ $d_{i} = x_{i} - a$	$u_i=\frac{d_i}{h}$ $u_{i} = h d ^{i}$	$f_i u_i$ $f_{i} u_{i}$
65–85	4	75	-60	-3	-12
85–105	5	95	-40	-2	-10
105–125	13	115	-20	-1	-13
125–145	20	135	0	0	0
145–165	14	155	20	1	14
165–185	8	175	40	2	16
185–205	4	195	60	3	12

$\sum f_i = 68,\quad \sum f_i u_i = 7$ $\sum f_{i} = 68, \sum f_{i} u_{i} = 7$

Now,

$\bar{X} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) = 135 + 20\left(\frac{7}{68}\right)$ $X ˉ = a + h (\sum f ^{i} \sum f ^{i} u ^{i}) = 135 + 20 (687)$

$= 135 + 2.0588 = 137.0588 \approx 137.05$ $= 135 + 2.0588 = 137.0588 \approx 137.05$

✅ Mean = 137.05 units

2) Mode

Highest frequency is 20, so modal class = 125–145

$h = 20$ $h = 20$
$l = 125$ $l = 125$
$f_1 = 20$ $f_{1} = 20$ (modal class frequency)
$f_0 = 13$ $f_{0} = 13$ (preceding class frequency)
$f_2 = 14$ $f_{2} = 14$ (succeeding class frequency)

$\text{Mode} = l + \left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h = 125 + \left(\frac{20-13}{2(20)-13-14}\right)20$ $Mode = l + (2 f ^{1} - f ^{0} - f ^{2} f ^{1} - f ^{0}) h = 125 + (2 ( 20 ) - 13 - 1420 - 13) 20$

$= 125 + \left(\frac{7}{40-27}\right)20 = 125 + \left(\frac{7}{13}\right)20$ $= 125 + (40 - 277) 20 = 125 + (137) 20$

$= 125 + 10.769 \approx 135.76$ $= 125 + 10.769 \approx 135.76$

✅ Mode = 135.76 units

3) Median

First we calculate cumulative frequency:

Monthly consumption (units)	$f$ $f$	$cf$ $c f$
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

$n = 68 \Rightarrow \frac{n}{2} = 34$ $n = 68 \Rightarrow 2 n = 34$

Cumulative frequency just greater than 34 is 42, so median class = 125–145

$l = 125$ $l = 125$
$h = 20$ $h = 20$
$f = 20$ $f = 20$
$cf = 22$ $c f = 22$ (cumulative frequency before median class)

$\text{Median} = l + \left(\frac{\frac{n}{2}-cf}{f}\right)h = 125 + \left(\frac{34-22}{20}\right)20$ $Median = l + (f 2 n - c f) h = 125 + (2034 - 22) 20$

$= 125 + 12 = 137$ $= 125 + 12 = 137$

✅ Median = 137 units

Final Answer + Comparison

Median = 137 units
Mode = 135.76 units
Mean = 137.05 units

Comparison & Interpretation:
Mean, median, and mode almost equal hain, iska matlab data almost symmetric / balanced hai. Is locality mein typical monthly consumption around 137 units ke aas-paas aata hai, aur sabse common usage class 125–145 units hai.

Q2. 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

Difficulty Level: Medium
Known: Median = 28.5, Total frequency = 60
Unknown: x and y

Reasoning

Median class is the class whose cumulative frequency is just greater than

$\frac{n}{2}$ $2 n$ .

$\text{Median} = l + \left(\frac{\frac{n}{2}-cf}{f}\right)h$ $Median = l + (f 2 n - c f) h$

Solution

Step 1: Cumulative frequency

Class Interval	Frequency	Cumulative Frequency
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	45 + x + y

Total

$n = 60$ $n = 60$

$45 + x + y = 60 \Rightarrow x + y = 15 \quad \text{...(i)}$ $45 + x + y = 60 \Rightarrow x + y = 15 ...(i)$

Step 2: Identify median class

$\frac{n}{2} = \frac{60}{2} = 30$ $2 n = 260 = 30$

Median = 28.5 lies in class 20–30, so median class is 20–30

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

Step 3: Apply median formula

$28.5 = 20 + \left(\frac{30-(5+x)}{20}\right)10$ $28.5 = 20 + (2030 - ( 5 + x )) 10$

$8.5 = \left(\frac{25-x}{20}\right)10$ $8.5 = (2025 - x) 10$

$8.5 = \frac{25-x}{2}$ $8.5 = 225 - x$

$17 = 25 - x \Rightarrow x = 8$ $17 = 25 - x \Rightarrow x = 8$

Now from (i):

$x + y = 15 \Rightarrow 8 + y = 15 \Rightarrow y = 7$ $x + y = 15 \Rightarrow 8 + y = 15 \Rightarrow y = 7$

✅ x = 8, y = 7

Q3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age (18 years onwards but less than 60 years).

Given (Less than type cumulative frequencies):
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

Difficulty Level: Medium
Known: Cumulative frequency data for 100 policy holders
Unknown: Median age

Reasoning

Convert “less than” cumulative data to class intervals and frequencies, then apply median formula.

Solution

Step 1: Convert to class intervals + frequencies

Age (years)	Cumulative Frequency (cf)	Frequency (f)
18–20	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

Total

$n=100$ $n = 100$

$\frac{n}{2} = 50$ $2 n = 50$

Cumulative frequency just greater than 50 is 78, so median class is 35–40

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

$\text{Median} = 35 + \left(\frac{50-45}{33}\right)5 = 35 + \frac{25}{33} = 35 + 0.7576 = 35.76$ $Median = 35 + (3350 - 45) 5 = 35 + 3325 = 35 + 0.7576 = 35.76$

✅ Median age = 35.76 years

Q4. Median length of leaves (convert to continuous classes)

Length (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

Difficulty Level: Medium
Known: Lengths of 40 leaves (nearest mm)
Unknown: Median length

Reasoning

Convert to continuous classes:
117.5–126.5, 126.5–135.5, ..., 171.5–180.5

Solution

Continuous class (mm)	f	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

$n=40 \Rightarrow \frac{n}{2}=20$ $n = 40 \Rightarrow 2 n = 20$
cf just greater than 20 is 29, so median class = 144.5–153.5

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

$\text{Median} = 144.5 + \left(\frac{20-17}{12}\right)9 = 144.5 + \left(\frac{3}{12}\right)9 = 144.5 + 2.25 = 146.75$ $Median = 144.5 + (1220 - 17) 9 = 144.5 + (123) 9 = 144.5 + 2.25 = 146.75$

✅ Median leaf length = 146.75 mm

Q5. Median lifetime of neon lamps

Lifetime (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

Total

$n = 400$ $n = 400$

Solution

Cumulative frequencies:

Class	f	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

$\frac{n}{2} = 200$ $2 n = 200$

cf just greater than 200 is 216, so median class = 3000–3500

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

$\text{Median} = 3000 + \left(\frac{200-130}{86}\right)500 = 3000 + \left(\frac{70}{86}\right)500$ $Median = 3000 + (86200 - 130) 500 = 3000 + (8670) 500$

$= 3000 + 406.98 = 3406.98$ $= 3000 + 406.98 = 3406.98$

✅ Median lifetime = 3406.98 hours

Q6. Surnames: Median, Mean, Mode (number of letters)

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

Total

$n = 100$ $n = 100$

(A) Median

Cumulative frequency:

Class	f	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

$\frac{n}{2} = 50$ $2 n = 50$

cf just greater than 50 is 76, so median class = 7–10

$l=7,\; h=3,\; f=40,\; cf=36$ $l = 7, h = 3, f = 40, c f = 36$

$\text{Median} = 7 + \left(\frac{50-36}{40}\right)3 = 7 + \left(\frac{14}{40}\right)3 = 7 + 1.05 = 8.05$ $Median = 7 + (4050 - 36) 3 = 7 + (4014) 3 = 7 + 1.05 = 8.05$

✅ Median = 8.05 letters

(B) Mean (Step-deviation)

Assumed mean

$a = 11.5$ $a = 11.5$ , class size

$h = 3$ $h = 3$

Class	f	$x_i$ $x_{i}$	$d_i$ $d_{i}$	$u_i$ $u_{i}$	$f u$ $f u$
1–4	6	2.5	-9	-3	-18
4–7	30	5.5	-6	-2	-60
7–10	40	8.5	-3	-1	-40
10–13	16	11.5	0	0	0
13–16	4	14.5	3	1	4
16–19	4	17.5	6	2	8

$\sum f = 100,\; \sum f u = -106$ $\sum f = 100, \sum f u = - 106$

$\bar{X} = a + h\left(\frac{\sum fu}{\sum f}\right) = 11.5 + 3\left(\frac{-106}{100}\right) = 11.5 - 3.18 = 8.32$ $X ˉ = a + h (\sum f \sum f u) = 11.5 + 3 (100-106) = 11.5 - 3.18 = 8.32$

✅ Mean = 8.32 letters

(C) Mode

Modal class = 7–10 (highest f = 40)

$l=7,\; h=3,\; f_1=40,\; f_0=30,\; f_2=16$ $l = 7, h = 3, f_{1} = 40, f_{0} = 30, f_{2} = 16$

$\text{Mode} = 7 + \left(\frac{40-30}{2(40)-30-16}\right)3 = 7 + \left(\frac{10}{80-46}\right)3$ $Mode = 7 + (2 ( 40 ) - 30 - 1640 - 30) 3 = 7 + (80 - 4610) 3$

$= 7 + \left(\frac{10}{34}\right)3 = 7 + 0.88 = 7.88$ $= 7 + (3410) 3 = 7 + 0.88 = 7.88$

✅ Mode = 7.88 letters

Final (Q6): Median = 8.05, Mean = 8.32, Mode = 7.88

Q7. Median weight of students

Weight (kg) – Number of students
40–45: 2
45–50: 3
50–55: 8
55–60: 6
60–65: 6
65–70: 3
70–75: 2

Total

$n = 30$ $n = 30$

Solution

Cumulative frequency:

Class	f	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

$\frac{n}{2} = 15$ $2 n = 15$

cf just greater than 15 is 19, so median class = 55–60

$l=55,\; h=5,\; f=6,\; cf=13$ $l = 55, h = 5, f = 6, c f = 13$

$\text{Median} = 55 + \left(\frac{15-13}{6}\right)5 = 55 + \left(\frac{2}{6}\right)5 = 55 + 1.6667 = 56.67$ $Median = 55 + (615 - 13) 5 = 55 + (62) 5 = 55 + 1.6667 = 56.67$

✅ Median weight = 56.67 kg

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

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

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

Class interval	Frequency
Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60	2 6 24 45 78 89 92 98 100

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

NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.3