Statistics 2-Mark Questions Class 10: CBSE PYQs with Answers

Given: Class intervals: 1–5, 5–9, 9–13, 13–17; frequencies: 4, 8, 7, 6.

To find: The mean.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: x̄ = $\frac{\text{235}}{\text{25}}$.

Calculation: Mean = $\frac{\text{47}}{\text{5}}$ = 9.4.

Answer: Mean = $\frac{\text{47}}{\text{5}}$ = 9.4

Given: Class intervals: 150–200, 200–250, 250–300, 300–350, 350–400; frequencies: 5, 6, 10, 2, 2.

To find: The mean in ₹.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: x̄ = $\frac{\text{6375}}{\text{25}}$.

Calculation: Mean = 255.

Answer: Mean = 255 rupees

Related method: Question 1.

Given: Class intervals: 0–40, 40–80, 80–120, 120–160, 160–200; frequencies: 12, 20, 35, 30, 23.

To find: The mean in runs.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: x̄ = $\frac{\text{13280}}{\text{120}}$.

Calculation: Mean = $\frac{\text{332}}{\text{3}}$ ≈ 110.67.

Answer: Mean = $\frac{\text{332}}{\text{3}}$ ≈ 110.67 runs

Keep the grouped mean as 110.67 runs to two decimal places. The official scheme also gives 111 runs when rounded to the nearest whole run.

Related method: Question 1.

Given: Class intervals: 0–20, 20–40, 40–60, 60–80, 80–100; frequencies: 8, 7, 12, 5, 3.

To find: The mode.

Identify the modal class: The greatest frequency is 12, so the modal class is 40–60. Thus l = 40, h = 20, f₀ = 7, f₁ = 12, f₂ = 5.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 40 + $\frac{\text{12 − 7}}{\text{2 × 12 − 7 − 5}}$ × 20.

Calculation: Mode = 40 + $\frac{\text{5}}{\text{12}}$ × 20.

Mode = 40 + $\frac{\text{25}}{\text{3}}$.

Answer: Mode = $\frac{\text{145}}{\text{3}}$ ≈ 48.33

Given: Class intervals: 10–20, 20–30, 30–40, 40–50, 50–60; frequencies: 15, 10, 12, 17, 4.

To find: The mode.

Identify the modal class: The greatest frequency is 17, so the modal class is 40–50. Thus l = 40, h = 10, f₀ = 12, f₁ = 17, f₂ = 4.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 40 + $\frac{\text{17 − 12}}{\text{2 × 17 − 12 − 4}}$ × 10.

Calculation: Mode = 40 + $\frac{\text{5}}{\text{18}}$ × 10.

Mode = 40 + $\frac{\text{25}}{\text{9}}$.

Answer: Mode = $\frac{\text{385}}{\text{9}}$ ≈ 42.78

Related method: Question 4.

Given: Class intervals: 20–30, 30–40, 40–50, 50–60, 60–70; frequencies: 25, 30, 45, 42, 35.

To find: The mode.

Identify the modal class: The greatest frequency is 45, so the modal class is 40–50. Thus l = 40, h = 10, f₀ = 30, f₁ = 45, f₂ = 42.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 40 + $\frac{\text{45 − 30}}{\text{2 × 45 − 30 − 42}}$ × 10.

Calculation: Mode = 40 + $\frac{\text{15}}{\text{18}}$ × 10.

Mode = 40 + $\frac{\text{25}}{\text{3}}$.

Answer: Mode = $\frac{\text{145}}{\text{3}}$ ≈ 48.33

Related method: Question 4.

Given: Class intervals: 100–110, 110–120, 120–130, 130–140, 140–150; frequencies: 5, 9, 8, 11, 7.

To find: The mode.

Identify the modal class: The greatest frequency is 11, so the modal class is 130–140. Thus l = 130, h = 10, f₀ = 8, f₁ = 11, f₂ = 7.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 130 + $\frac{\text{11 − 8}}{\text{2 × 11 − 8 − 7}}$ × 10.

Calculation: Mode = 130 + $\frac{\text{3}}{\text{7}}$ × 10.

Mode = 130 + $\frac{\text{30}}{\text{7}}$.

Answer: Mode = $\frac{\text{940}}{\text{7}}$ ≈ 134.29

Related method: Question 4.

Given: Class intervals: 1–3, 3–5, 5–7, 7–9, 9–11, 11–13; frequencies: 20, 45, 80, 55, 40, 12.

To find: The mode in hectares.

Identify the modal class: The greatest frequency is 80, so the modal class is 5–7. Thus l = 5, h = 2, f₀ = 45, f₁ = 80, f₂ = 55.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 5 + $\frac{\text{80 − 45}}{\text{2 × 80 − 45 − 55}}$ × 2.

Calculation: Mode = 5 + $\frac{\text{35}}{\text{60}}$ × 2.

Mode = 5 + $\frac{\text{7}}{\text{6}}$.

Answer: Mode = $\frac{\text{37}}{\text{6}}$ ≈ 6.17 hectares

Related method: Question 4.

Given: Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60; frequencies: 3, 9, 15, 30, 18, 5.

To find: The mode.

Identify the modal class: The greatest frequency is 30, so the modal class is 30–40. Thus l = 30, h = 10, f₀ = 15, f₁ = 30, f₂ = 18.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 30 + $\frac{\text{30 − 15}}{\text{2 × 30 − 15 − 18}}$ × 10.

Calculation: Mode = 30 + $\frac{\text{15}}{\text{27}}$ × 10.

Mode = 30 + $\frac{\text{50}}{\text{9}}$.

Answer: Mode = $\frac{\text{320}}{\text{9}}$ ≈ 35.56

Related method: Question 4.

Given: Class intervals: 15–25, 25–35, 35–45, 45–55, 55–65, 65–75; frequencies: 6, 11, 22, 23, 14, 5.

To find: The mode.

Identify the modal class: The greatest frequency is 23, so the modal class is 45–55. Thus l = 45, h = 10, f₀ = 22, f₁ = 23, f₂ = 14.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 45 + $\frac{\text{23 − 22}}{\text{2 × 23 − 22 − 14}}$ × 10.

Calculation: Mode = 45 + $\frac{\text{1}}{\text{10}}$ × 10.

Mode = 45 + 1.

Answer: Mode = 46

Related method: Question 4.

Given: Class intervals: 25–30, 30–35, 35–40, 40–45, 45–50; frequencies: 12, 5, 14, 8, 9.

To find: The mode.

Identify the modal class: The greatest frequency is 14, so the modal class is 35–40. Thus l = 35, h = 5, f₀ = 5, f₁ = 14, f₂ = 8.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 35 + $\frac{\text{14 − 5}}{\text{2 × 14 − 5 − 8}}$ × 5.

Calculation: Mode = 35 + $\frac{\text{9}}{\text{15}}$ × 5.

Mode = 35 + 3.

Answer: Mode = 38

Related method: Question 4.

Given: Class intervals: 0–20, 20–40, 40–60, 60–80, 80–100; frequencies: 20, 29, 53, 61, 37.

To find: The mode.

Identify the modal class: The greatest frequency is 61, so the modal class is 60–80. Thus l = 60, h = 20, f₀ = 53, f₁ = 61, f₂ = 37.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 60 + $\frac{\text{61 − 53}}{\text{2 × 61 − 53 − 37}}$ × 20.

Calculation: Mode = 60 + $\frac{\text{8}}{\text{32}}$ × 20.

Mode = 60 + 5.

Answer: Mode = 65

Related method: Question 4.

Given: Class intervals: 25–35, 35–45, 45–55, 55–65, 65–75; frequencies: 6, 10, 8, 12, 4.

To find: The mode.

Identify the modal class: The greatest frequency is 12, so the modal class is 55–65. Thus l = 55, h = 10, f₀ = 8, f₁ = 12, f₂ = 4.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 55 + $\frac{\text{12 − 8}}{\text{2 × 12 − 8 − 4}}$ × 10.

Calculation: Mode = 55 + $\frac{\text{4}}{\text{12}}$ × 10.

Mode = 55 + $\frac{\text{10}}{\text{3}}$.

Answer: Mode = $\frac{\text{175}}{\text{3}}$ ≈ 58.33

Related method: Question 4.

Given: Class intervals: 16–18, 18–20, 20–22, 22–24, 24–26; frequencies: 50, 78, 46, 28, 23.

To find: The mode in years.

Identify the modal class: The greatest frequency is 78, so the modal class is 18–20. Thus l = 18, h = 2, f₀ = 50, f₁ = 78, f₂ = 46.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 18 + $\frac{\text{78 − 50}}{\text{2 × 78 − 50 − 46}}$ × 2.

Calculation: Mode = 18 + $\frac{\text{28}}{\text{60}}$ × 2.

Mode = 18 + $\frac{\text{14}}{\text{15}}$.

Answer: Mode = $\frac{\text{284}}{\text{15}}$ ≈ 18.93 years

The exact grouped estimate is 284/15 years ≈ 18.93 years.

Related method: Question 4.

Given: Class intervals: 5–10, 10–15, 15–20, 20–25, 25–30, 30–35; frequencies: 5, 6, 15, 10, 5, 4.

To find: The mode.

Identify the modal class: The greatest frequency is 15, so the modal class is 15–20. Thus l = 15, h = 5, f₀ = 6, f₁ = 15, f₂ = 10.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 15 + $\frac{\text{15 − 6}}{\text{2 × 15 − 6 − 10}}$ × 5.

Calculation: Mode = 15 + $\frac{\text{9}}{\text{14}}$ × 5.

Mode = 15 + $\frac{\text{45}}{\text{14}}$.

Answer: Mode = $\frac{\text{255}}{\text{14}}$ ≈ 18.21

Related method: Question 4.

Given: Class intervals: 1–3, 3–5, 5–7, 7–9, 9–11; frequencies: 7, 8, 2, 2, 1.

To find: The mode in members.

Identify the modal class: The greatest frequency is 8, so the modal class is 3–5. Thus l = 3, h = 2, f₀ = 7, f₁ = 8, f₂ = 2.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: Mode = 3 + $\frac{\text{8 − 7}}{\text{2 × 8 − 7 − 2}}$ × 2.

Calculation: Mode = 3 + $\frac{\text{1}}{\text{7}}$ × 2.

Mode = 3 + $\frac{\text{2}}{\text{7}}$.

Answer: Mode = $\frac{\text{23}}{\text{7}}$ ≈ 3.29 members

The formula gives a grouped-data estimate of the mode. A fractional estimate is possible even though individual family sizes are whole numbers.

Related method: Question 4.

Given: Class intervals: 0–20, 20–40, 40–60, 60–80, 80–100; frequencies: 11, 22, 19, 13, 7.

To find: The modal class and the median class.

Rule / formula: The modal class has the greatest frequency. For the median class, calculate N = Σfᵢ and N/2; take the class whose cumulative frequency first exceeds N/2.

Substitution: N = 72; N/2 = 36.

Calculation: The cumulative frequency is 33 before 40–60 and 52 at its end. Hence the median class is 40–60.

The greatest frequency is 22, so the modal class is 20–40.

Answer: Modal class: 20–40; median class: 40–60

Given: Class intervals: 50–55, 55–60, 60–65, 65–70, 70–75, 75–80; frequencies: 10, 15, 8, 13, 9, 5.

To find: The lower limit of the modal class plus the upper limit of the median class.

Rule / formula: The modal class has the greatest frequency. For the median class, calculate N = Σfᵢ and N/2; take the class whose cumulative frequency first exceeds N/2.

Substitution: N = 60; N/2 = 30.

Calculation: The cumulative frequency is 25 before 60–65 and 33 at its end. Hence the median class is 60–65.

The greatest frequency is 15, so the modal class is 55–60.

Required sum = 55 + 65.

Answer: Required sum = 120

Related method: Question 17.

Given: Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60; frequencies: 5, 8, 20, 15, 7, 5.

To find: The median in marks.

Rule / formula: Median = l + $\frac{\text{N/2 − cf}}{\text{f}}$ × h, where l is the lower limit of the median class, cf is the cumulative frequency before it, f is its frequency, h is its width and N is the total frequency.

Substitution: N = 60; N/2 = 30.

Calculation: The cumulative frequency is 13 before 20–30 and 33 at its end. Hence the median class is 20–30.

l = 20, cf = 13, f = 20, h = 10.

Median = 20 + $\frac{\text{30 − 13}}{\text{20}}$ × 10.

Median = 20 + $\frac{\text{17}}{\text{2}}$.

Answer: Median = $\frac{\text{57}}{\text{2}}$ = 28.5 marks

Related method: Question 17.

Given: Class intervals: 1–3, 3–5, 5–7, 7–9, 9–11; frequencies: 5, 4, 18, 16, 17.

To find: The lower limit of the modal class plus the upper limit of the median class.

Rule / formula: The modal class has the greatest frequency. For the median class, calculate N = Σfᵢ and N/2; take the class whose cumulative frequency first exceeds N/2.

Substitution: N = 60; N/2 = 30.

Calculation: The cumulative frequency is 27 before 7–9 and 43 at its end. Hence the median class is 7–9.

The greatest frequency is 18, so the modal class is 5–7.

Required sum = 5 + 9.

Answer: Required sum = 14

Related method: Question 17.

Given: Class intervals: 0–5, 5–10, 10–15, 15–20, 20–25; frequencies: 2, 9, 6, 7, 1.

To find: The lower limit of the modal class plus the upper limit of the median class.

Rule / formula: The modal class has the greatest frequency. For the median class, calculate N = Σfᵢ and N/2; take the class whose cumulative frequency first exceeds N/2.

Substitution: N = 25; N/2 = $\frac{\text{25}}{\text{2}}$.

Calculation: The cumulative frequency is 11 before 10–15 and 17 at its end. Hence the median class is 10–15.

The greatest frequency is 9, so the modal class is 5–10.

Required sum = 5 + 15.

Answer: Required sum = 20

Related method: Question 17.

Given: Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50; frequencies: 5, 7, a, 5, b. The cumulative frequencies are 5, c, 18, d and 30, respectively.

To find: The missing entries a, b, c and d.

Rule / formula: Cumulative frequency at a class = previous cumulative frequency + its frequency. Therefore a class frequency = current cumulative frequency − previous cumulative frequency.

Substitution and calculation: c = 5 + 7 = 12.

a = 18 − c = 18 − 12 = 6.

d = 18 + 5 = 23.

b = 30 − d = 30 − 23 = 7.

Check: 5 + 7 + 6 + 5 + 7 = 30, the final cumulative frequency.

Answer: a = 6; b = 7; c = 12; d = 23

Related method: Question 17.

Given: Class intervals: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90; frequencies: 10, 7, x, 15, 10, 12. Mode = 55.

To find: The missing frequency x.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Identify the modal class: The given mode 55 lies in 45–60. Thus l = 45, h = 15, f₁ = 15, f₀ = x and f₂ = 10.

Substitution: 55 = 45 + $\frac{\text{15 − x}}{\text{30 − x − 10}}$ × 15.

Calculation: 10 = $\frac{\text{15(15 − x)}}{\text{20 − x}}$.

10(20 − x) = 15(15 − x).

200 − 10x = 225 − 15x.

5x = 25.

Check: x = 5 is non-negative and leaves 45–60 as the modal class.

Answer: x = 5

Given: Mode = 75; modal class = 65–80; preceding frequency f₀ = 6; succeeding frequency f₂ = 8. Hence l = 65 and h = 80 − 65 = 15.

To find: The modal-class frequency f₁.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Substitution: 75 = 65 + $\frac{\text{f₁ − 6}}{\text{2f₁ − 6 − 8}}$ × 15.

Calculation: 10(2f₁ − 14) = 15(f₁ − 6).

20f₁ − 140 = 15f₁ − 90.

5f₁ = 50.

Check: f₁ = 10 exceeds both neighbouring frequencies, 6 and 8.

Answer: Frequency of the modal class = 10

Related method: Question 23.

Given: Class intervals: 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25; frequencies: 3, 6, 9, 13, f, 5, 4. Mean = 18.

To find: The missing frequency f.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: 18 = $\frac{\text{20f + 704}}{\text{f + 40}}$.

Calculation: 18(f + 40) = 20f + 704.

18f + 720 = 20f + 704.

2f = 16.

f = $\frac{\text{16}}{\text{2}}$ = 8.

Check: With f = 8, Σfᵢ = 48 and Σfᵢxᵢ = 864; their ratio is 18.

Answer: f = 8

Related method: Question 23.

Given: Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50; frequencies: 4, x, 20, y, 6. Mode = 24. Total frequency N = 50.

To find: The missing frequencies x and y.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Total-frequency equation: 4 + x + 20 + y + 6 = 50.

x + y = 20, so y = 20 − x.

Identify the modal class: The given mode 24 lies in 20–30. Thus l = 20, h = 10, f₁ = 20, f₀ = x and f₂ = y.

Substitution: 24 = 20 + $\frac{\text{20 − x}}{\text{40 − x − y}}$ × 10.

24 = 20 + $\frac{\text{20 − x}}{\text{20}}$ × 10, since x + y = 20.

Calculation: 4 = $\frac{\text{20 − x}}{\text{2}}$.

8 = 20 − x.

x = 12.

y = 20 − 12 = 8.

Check: 4 + 12 + 20 + 8 + 6 = 50, and the greatest frequency is 20.

Answer: x = 12; y = 8

Related method: Question 23.

Given: Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50; frequencies: 15, 20, 35, k, 10. Mean = 24.

To find: The missing frequency k.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: 24 = $\frac{\text{35k + 1700}}{\text{k + 80}}$.

Calculation: 24(k + 80) = 35k + 1700.

24k + 1920 = 35k + 1700.

11k = 220.

k = $\frac{\text{220}}{\text{11}}$ = 20.

Check: With k = 20, Σfᵢ = 100 and Σfᵢxᵢ = 2400; their ratio is 24.

Answer: k = 20

Related method: Question 23.

Given: Class intervals: 0–100, 100–200, 200–300, 300–400, 400–500; frequencies: 140, 230, 270, x, 150. Mode = 240.

To find: The missing frequency x.

Formula: Mode = l + $\frac{\text{f₁ − f₀}}{\text{2f₁ − f₀ − f₂}}$ × h. Here l is the lower limit of the modal class, h its width, f₁ its frequency, f₀ the preceding frequency and f₂ the succeeding frequency.

Identify the modal class: The given mode 240 lies in 200–300. Thus l = 200, h = 100, f₁ = 270, f₀ = 230 and f₂ = x.

Substitution: 240 = 200 + $\frac{\text{270 − 230}}{\text{540 − 230 − x}}$ × 100.

Calculation: 40 = $\frac{\text{4000}}{\text{310 − x}}$.

40(310 − x) = 4000.

310 − x = 100.

x = 210.

Check: 210 is non-negative and less than 270; 200–300 remains the modal class.

Answer: x = 210

Related method: Question 23.

Given: Class intervals: 0–4, 4–8, 8–12, 12–16, 16–20; frequencies: 3, p, 5, 8, 2. Mean = 10.8.

To find: The missing frequency p.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: 10.8 = $\frac{\text{6p + 204}}{\text{p + 18}}$.

Calculation: 10.8(p + 18) = 6p + 204.

10.8p + 194.4 = 6p + 204.

4.8p = 9.6.

p = $\frac{\text{9.6}}{\text{4.8}}$ = 2.

Check: With p = 2, Σfᵢ = 20 and Σfᵢxᵢ = 216; their ratio is 10.8.

Answer: p = 2

Related method: Question 23.

Given: Class intervals: 6–10, 10–14, 14–18, 18–22, 22–26; frequencies: 4, 6, k, 5, 3. Mean = 15.6.

To find: The missing frequency k.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: 15.6 = $\frac{\text{16k + 276}}{\text{k + 18}}$.

Calculation: 15.6(k + 18) = 16k + 276.

15.6k + 280.8 = 16k + 276.

0.4k = 4.8.

k = $\frac{\text{4.8}}{\text{0.4}}$ = 12.

Check: With k = 12, Σfᵢ = 30 and Σfᵢxᵢ = 468; their ratio is 15.6.

Answer: k = 12

Related method: Question 23.

Given: Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50; frequencies: 5, 18, 15, f, 6. Mean = 25.

To find: The missing frequency f.

Formula: Class mark xᵢ = $\frac{\text{lower class limit + upper class limit}}{\text{2}}$; mean x̄ = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$, where fᵢ is the frequency of class i and Σ means the sum over all classes.

Substitution: 25 = $\frac{\text{35f + 940}}{\text{f + 44}}$.

Calculation: 25(f + 44) = 35f + 940.

25f + 1100 = 35f + 940.

10f = 160.

f = $\frac{\text{160}}{\text{10}}$ = 16.

Check: With f = 16, Σfᵢ = 60 and Σfᵢxᵢ = 1500; their ratio is 25.

Answer: f = 16

Related method: Question 23.

Given: Mean = 169; mode = 175.

To find: The median using the stated empirical relationship.

Formula: Mode = 3 × Median − 2 × Mean. This is the empirical relationship requested in the question.

Substitution: 175 = 3 × Median − 2 × 169.

Calculation: 3 × Median = 175 + 338.

3 × Median = 513.

Median = $\frac{\text{513}}{\text{3}}$.

Answer: Median = 171

Given: Mean = 10.5; mode = 12.4.

To find: The median using the stated empirical relationship.

Formula: Mode = 3 × Median − 2 × Mean. This is the empirical relationship requested in the question.

Substitution: 12.4 = 3 × Median − 2 × 10.5.

Calculation: 3 × Median = 12.4 + 21.

3 × Median = 33.4.

Median = $\frac{\text{33.4}}{\text{3}}$.

Answer: Median = $\frac{\text{167}}{\text{15}}$ ≈ 11.13

Related method: Question 32.