NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.2

Q1. Ages of patients admitted in a hospital during a year (Mean + Mode + Interpretation)

Difficulty Level: Moderate

Given (Known):
Age (in years): 5–15, 15–25, 25–35, 35–45, 45–55, 55–65
Number of Patients (f): 6, 11, 21, 23, 14, 5

To Find (Unknown): Mean and Mode, and compare/interpret both.

Method (Reasoning):

• Mean: Direct method (using class marks).

• Mode: Grouped data mode formula (modal class = highest frequency).

Solution

(A) Mean

Class mark: xᵢ = (lower + upper)/2

∑f = 6+11+21+23+14+5 = 80
∑(f·x) = 60+220+630+920+700+300 = 2830

Mean, x̄ = ∑(f·x) / ∑f = 2830/80 = 35.375 ≈ 35.37 years

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(B) Mode

Highest frequency = 23 → modal class = 35–45

Mode formula (grouped):
Mode = l + \(f₁ − f₀\) / \(2f₁ − f₀ − f₂\) × h

Here:
l = 35, h = 10
f₁ = 23 (modal class)
f₀ = 21 (previous class 25–35)
f₂ = 14 (next class 45–55)

Mode = 35 +

$(23−21)/(2×23−21−14)$(23−21)/(2×23−21−14)×10
= 35 +

$2/(46−35)$2/(46−35)×10
= 35 + (2/11)×10
= 35 + 1.818…
= 36.8 years (approx.)

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(C) Compare & Interpret

• Mean = 35.37 years → average age of patients.

• Mode = 36.8 years → most common age (highest concentration) lies around 36–37 years (modal class 35–45).

• Since mean and mode are close, data is not highly skewed; most patients fall near the mid-30s to early-40s.

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Q2. Observed lifetimes (in hours) of 225 electric components (Find Mode)

Difficulty Level: Moderate

Given (Known):
Lifetime (hours): 0–20, 20–40, 40–60, 60–80, 80–100, 100–120
Frequency: 10, 35, 52, 61, 38, 29

To Find (Unknown): Modal lifetime.

Method (Reasoning):
Modal class = highest frequency, then use grouped mode formula.

Solution

Highest frequency = 61 → modal class = 60–80

Mode = l + \(f₁ − f₀\) / \(2f₁ − f₀ − f₂\) × h

l = 60, h = 20
f₁ = 61
f₀ = 52 (40–60)
f₂ = 38 (80–100)

Mode = 60 +

$(61−52)/(2×61−52−38)$(61−52)/(2×61−52−38)×20
= 60 +

$9/(122−90)$9/(122−90)×20
= 60 + (9/32)×20
= 60 + 5.625
= 65.625 hours

So, the modal lifetime ≈ 65.625 hours.

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Q3. Monthly household expenditure of 200 families (Find Mode + Mean)

Difficulty Level: Moderate

Given (Known):

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

To Find (Unknown): Mean and Mode.

Method (Reasoning):

• Mean: Step-deviation method (class width same = 500).

• Mode: grouped mode formula.

Solution

(A) Mean (Step-deviation)

Class size h = 500
Assumed mean a = 2750 (mid class 2500–3000)

Class marks xᵢ: 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750
dᵢ = xᵢ − a ; uᵢ = dᵢ / h

∑f = 200
∑(f·u) = -72-80-33+0+30+44+48+28 = -35

Mean formula:
x̄ = a + h × (∑f·u / ∑f)
= 2750 + 500 × (-35/200)
= 2750 + 500 × (-0.175)
= 2750 − 87.5
= 2662.5

Mean monthly expenditure = ₹ 2662.50

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(B) Mode

Highest frequency = 40 → modal class = 1500–2000

l = 1500, h = 500
f₁ = 40
f₀ = 24 (1000–1500)
f₂ = 33 (2000–2500)

Mode = 1500 +

$(40−24)/(2×40−24−33)$(40−24)/(2×40−24−33)×500
= 1500 +

$16/(80−57)$16/(80−57)×500
= 1500 + (16/23)×500
= 1500 + 347.826…
= ₹ 1847.83 (approx.)

So, Mode ≈ ₹ 1847.83 and Mean = ₹ 2662.50.

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Q4. State-wise teacher-student ratio (Find Mode + Mean + Interpretation)

Difficulty Level: Moderate

Given (Known):

Students per teacher: 15–20, 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55
No. of states/UT: 3, 8, 9, 10, 3, 0, 0, 2

To Find (Unknown): Mean and Mode, and interpret.

Method (Reasoning):

• Mean: direct method (class marks).

• Mode: grouped mode formula.

Solution

(A) Mean

Class marks xᵢ: 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5

Compute f·x:

15–20: 3×17.5 = 52.5
20–25: 8×22.5 = 180
25–30: 9×27.5 = 247.5
30–35: 10×32.5 = 325
35–40: 3×37.5 = 112.5
40–45: 0×42.5 = 0
45–50: 0×47.5 = 0
50–55: 2×52.5 = 105

∑f = 35
∑(f·x) = 52.5+180+247.5+325+112.5+0+0+105 = 1022.5

Mean = 1022.5 / 35 = 29.214… ≈ 29.2

Mean ratio ≈ 29.2 students per teacher

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(B) Mode

Highest frequency = 10 → modal class = 30–35

l = 30, h = 5
f₁ = 10
f₀ = 9 (25–30)
f₂ = 3 (35–40)

Mode = 30 +

$(10−9)/(2×10−9−3)$(10−9)/(2×10−9−3)×5
= 30 +

$1/(20−12)$1/(20−12)×5
= 30 + (1/8)×5
= 30 + 0.625
= 30.625 ≈ 30.6

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Interpretation:

• Mean ≈ 29.2 shows the average student-teacher ratio across states/UTs.

• Mode ≈ 30.6 shows the most common ratio band is around 30–35.

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Q5. Runs scored by top batsmen in ODI cricket (Find Mode)

Difficulty Level: Moderate

Given (Known):

Runs scored (ODI):
3000–4000: 4
4000–5000: 18
5000–6000: 9
6000–7000: 7
7000–8000: 6
8000–9000: 3
9000–10000: 1
10000–11000: 1

To Find (Unknown): Mode.

Method (Reasoning):
Modal class = highest frequency, then mode formula.

Solution

Highest frequency = 18 → modal class = 4000–5000

l = 4000, h = 1000
f₁ = 18
f₀ = 4 (3000–4000)
f₂ = 9 (5000–6000)

Mode = 4000 +

$(18−4)/(2×18−4−9)$(18−4)/(2×18−4−9)×1000
= 4000 +

$14/(36−13)$14/(36−13)×1000
= 4000 + (14/23)×1000
= 4000 + 608.695…
= 4608.7 (approx.)

Mode ≈ 4608.7 runs

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Q6. Cars passing through a spot (Find Mode)

Difficulty Level: Moderate

Given (Known):
Number of cars (per 3 minutes): 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

To Find (Unknown): Mode.

Method (Reasoning):
Modal class = highest frequency, then grouped mode formula.

Solution

Highest frequency = 20 → modal class = 40–50

l = 40, h = 10
f₁ = 20
f₀ = 12 (30–40)
f₂ = 11 (50–60)

Mode = 40 +

$(20−12)/(2×20−12−11)$(20−12)/(2×20−12−11)×10
= 40 +

$8/(40−23)$8/(40−23)×10
= 40 + (8/17)×10
= 40 + 4.705…
= 44.7 (approx.)

So, Mode ≈ 44.7 cars (per 3-minute period).