Tag: Exercise 13.2 Chapter 13 Class 10th Maths Statistics NCERT Solutions

Exercise-13.3, Class 10th, Maths, Chapter 13, NCERT
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.

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

Solution
1. Compute class-marks $x$ and $f$ :
Class f class-mark x f·x

65–85 4 75 300

85–105 5 95 475

105–125 13 115 1495

125–145 20 135 2700

145–165 14 155 2170

165–185 8 175 1400

185–205 4 195 780 --------------------------------- Total 68 Σf x = 9320
1. Mean
$\overset{ˉ}{x} = \frac{Σ f x}{Σ f} = \frac{9320}{68} = 137.0588 \approx 137.06 units$
1. Median
$N = 68$ . $N / 2 = 34.Cumulative frequencies:$

65–85: 4

85–105: 9

105–125: 22

125–145: 42

← median class (since 34 lies in this class)

Median class: $125 - 145$ .
$l = 125, h = 20, f = 20, c.f. before = 22$

Median formula:

$Median = l + \frac{\frac{N}{2} - c.f. before}{f} \times h = 125 + \frac{34 - 22}{20} \times 20$

$= 125 + 12 = 137.0 units$
1. Mode
Modal class = class with largest frequency = $125 - 145$ with $f_{1} = 20$
$f_{0} = 13$ (preceding), $f_{2} = 14$ (succeeding), $l = 125, h = 20$

$Mode = l + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h = 125 + \frac{20 - 13}{40 - 13 - 14} \times 20$

$= 125 + \frac{7}{13} \times 20 = 125 + 10.7692 \approx 135.77 units$

Comparison / Interpretation:
Mean $\approx 137.06$ , Median $= 137.0$ , Mode $\approx 135.77$ — all three measures are very close, indicating a fairly symmetric distribution around about 136–137 units.

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

Solution

Median is $28.5$ → it lies in class $20 - 30$ . Use median formula:

Here $N = 60$ , so $N / 2 = 30$ . Median class: $20 - 30$ with $l = 20, h = 10, f = 20$ . c.f. before = frequency up to previous class = $5 + x$ .

Median:

$28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10 = 20 + \frac{25 - x}{20} \times 10 = 20 + \frac{25 - x}{2}$

So

$28.5 = 20 + \frac{25 - x}{2} \Rightarrow \frac{25 - x}{2} = 8.5 \Rightarrow 25 - x = 17 \Rightarrow x = 8.$

Now total frequencies:

$5 + x + 20 + 15 + y + 5 = 60$

Substitute $x = 8$ :

$5 + 8 + 20 + 15 + y + 5 = 60 \Rightarrow 53 + y = 60 \Rightarrow y = 7.$

Answer: $x = 8, y = 7$

Q3

A life insurance agent found the following data for distribution of ages of 100 policy holders (cumulative ‘below’ form). 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 …)
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

Solution

First convert to class frequencies (by differences):

20–25: 6 − 2 = 4

25–30: 24 − 6 = 18

30–35: 45 − 24 = 21

35–40: 78 − 45 = 33

40–45: 89 − 78 = 11

45–50: 92 − 89 = 3

50–55: 98 − 92 = 6

55–60:100 − 98 = 2 (And below 20 group: 2)

Total $N = 100$ . Median position $= N / 2 = 50$ . Cumulative frequencies:

Below20: 2

20–25: 2+4 = 6

25–30: 6 +18 = 24

30–35: 24 +21 = 45

35–40: 45 +33 = 78

← median class (since 50th lies here)

Median class = $35 - 40$ . Use: $l = 35, h = 5, f = 33, c.f. before = 45$

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

$= 35 + \frac{25}{33} \approx 35 + 0.7576 = 35.76 years (approx)$

Q4

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre; the data:

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

Find the median length of the leaves.
(Hint: convert to continuous classes: 117.5–126.5, 126.5–135.5, …, 171.5–180.5.)

Solution

Use continuous class limits (class-width $h = 9$ ):

117.5–126.5 : 3 (cf = 3)

126.5–135.5 : 5 (cf = 8)

135.5–144.5 : 9 (cf = 17)

144.5–153.5 :12 (cf = 29)

← median class (since N/2 = 20)

153.5–162.5 : 5 (cf = 34)

162.5–171.5 : 4 (cf = 38)

171.5–180.5 : 2 (cf = 40)

$N = 40, N / 2 = 20.$ Median class: $144.5 - 153.5$
$l$ c.f. before $= 17$

$Median = l + \frac{\frac{N}{2} - c.f. before}{f} \times h = 144.5 + \frac{20 - 17}{12} \times 9$

$= 144.5 + \frac{3}{12} \times 9 = 144.5 + 2.25 = 146.75 mm$

Q5

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

Life time (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

Solution

Total $N = 400, N / 2 = 200$ . Cumulative frequencies:

1500–2000 : 14

2000–2500 : 14+56 = 70

2500–3000 : 70+60 = 130

3000–3500 : 130+86 = 216

← median class (since 200th lies here)

Median class = $3000 - 3500$ $l = 3000, h = 500, f = 86, c.f. before = 130$

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

$= 3000 + 0.8139535 \times 500$

$= 3000 + 406.9767 \approx 3406.98 hours$

Answer can be rounded: $3407 hours (approx)$

Q6

100 surnames were picked; frequency distribution of 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

Determine the median number of letters and find the mean number of letters.

Solution
1. Median
$N = 100, N / 2 = 50.$ Cumulative frequencies:

1–4 : 6

4–7 : 6+30 = 36

7–10: 36+40 = 76

← median class (50th lies here)

Use continuous limits with class-width $h = 3$ (e.g. 0.5–4.5, 4.5–7.5, 7.5–10.5 …). Median class interval in continuous form = $7.5 - 10.5$ So $l = 7.5, h = 3, f = 40,$ c.f. before = 36.

$Median = 7.5 + \frac{50 - 36}{40} \times 3 = 7.5 + \frac{14}{40} \times 3 = 7.5 + 1.05 = 8.55 letters (approx)$
1. Mean
Class-marks $x$ : 2.5, 5.5, 8.5, 11.5, 14.5, 17.5
Frequencies $f$ : 6, 30, 40, 16, 4, 4

Compute $f \cdot x$ :

2.5·6 = 15

5.5·30 = 165 (total 180)

8.5·40 = 340 (total 520)

11.5·16 = 184 (total 704)

14.5·4 = 58 (total 762)

17.5·4 = 70 (total 832)

Σf x = 832

$\overset{ˉ}{x} = \frac{Σ f x}{Σ f} = \frac{832}{100} = 8.32 letters$
October 12, 2025
Exercise-13.2, Class 10th, Maths, Chapter 13, NCERT

Question 1

The following table shows the ages of the patients admitted in a hospital during a year:

Age (years) 5–15 15–25 25–35 35–45 45–55 55–65

No. of patients 6 11 21 23 14 5

Find the mean and mode of the data. Compare the two measures.

Solution:

Total frequency (Σf) = 6 + 11 + 21 + 23 + 14 + 5 = 80

Class marks (x): 10, 20, 30, 40, 50, 60

f × x = 60 + 220 + 630 + 920 + 700 + 300 = 2830

Mean = Σ(fx) / Σf = 2830 / 80 = 35.38 years

Modal class = 35–45
l = 35, h = 10, f₁ = 23, f₀ = 21, f₂ = 14

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 35 + [(23 − 21) / (46 − 21 − 14)] × 10
= 35 + (2 / 11) × 10 = 36.82 years

Interpretation:
Mean = 35.38 years, Mode = 36.82 years.
Both are close, showing most patients are around 35–37 years old.

Question 2

Lifetimes (in hours) of 225 electrical components:

Lifetime (hours) 0–20 20–40 40–60 60–80 80–100 100–120

Frequency 10 35 52 61 38 29

Find the modal lifetime.

Solution:

Modal class = 60–80
l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 60 + [(61 − 52) / (122 − 52 − 38)] × 20
= 60 + (9 / 32) × 20
= 60 + 5.625 = 65.63 hours

Question 3

Monthly household expenditure of 200 families:

Expenditure (₹) 1000–1500 1500–2000 2000–2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000

No. of families 24 40 33 28 30 22 16 7

Find the mode and mean monthly expenditure.

Solution:

Modal class = 1500–2000
l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 1500 + [(40 − 24) / (80 − 24 − 33)] × 500
= 1500 + (16 / 23) × 500 = 1500 + 347.83 = ₹1847.83

Class marks (x): 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750

Σ(fx) = 532500, Σf = 200

Mean = Σ(fx)/Σf = 532500 / 200 = ₹2662.50

Interpretation:
Mean (₹2662.50) > Mode (₹1847.83), showing the data is right-skewed (some families spend much more).

Question 4

Teacher–student ratio (students per teacher):

Students per teacher 15–20 20–25 25–30 30–35 35–40 40–45 45–50 50–55

No. of states/U.T. 3 8 9 10 3 0 0 2

Find the mode and mean of this data.

Solution:

Modal class = 30–35
l = 30, h = 5, f₁ = 10, f₀ = 9, f₂ = 3

Mode = 30 + [(10 − 9) / (20 − 9 − 3)] × 5
= 30 + (1 / 8) × 5 = 30.63

Class marks (x): 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5

Σ(fx) = 1022.5, Σf = 35

Mean = 1022.5 / 35 = 29.21

Interpretation:
Mean = 29.21, Mode = 30.63 → Both indicate roughly 30 students per teacher.

Question 5

Runs scored by top batsmen in one-day internationals:

Runs scored 3000–4000 4000–5000 5000–6000 6000–7000 7000–8000 8000–9000 9000–10000 10000–11000

No. of batsmen 4 18 9 7 6 3 1 1

Find the mode of the data.

Solution:

Modal class = 4000–5000
l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9

Mode = 4000 + [(18 − 4) / (36 − 4 − 9)] × 1000
= 4000 + (14 / 23) × 1000
= 4000 + 608.7 = 4608.7 runs

Question 6

Number of cars passing a spot during 100 three-minute intervals:

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

Find the mode.

Solution:

Modal class = 40–50
l = 40, h = 10, f₁ = 20, f₀ = 12, f₂ = 11

Mode = 40 + [(20 − 12) / (40 − 12 − 11)] × 10
= 40 + (8 / 17) × 10
= 40 + 4.71 = 44.71 cars

Final Answers Summary

Question Mean Mode

1 35.38 36.82

2 — 65.63

3 2662.50 1847.83

4 29.21 30.63

5 — 4608.7

6 — 44.71

October 12, 2025

Question	Mean	Mode
1	35.38	36.82
2	—	65.63
3	2662.50	1847.83
4	29.21	30.63
5	—	4608.7
6	—	44.71

Tag: Exercise 13.2 Chapter 13 Class 10th Maths Statistics NCERT Solutions

Exercise-13.3, Class 10th, Maths, Chapter 13, NCERT

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Q4

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Q6

Exercise-13.2, Class 10th, Maths, Chapter 13, NCERT

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Question 6

Final Answers Summary