Statistics Case Study Questions Class 10: CBSE PYQs with Answers

Median = 58 years.

n = 140; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 70.

The first cumulative frequency greater than 70 is 98.

Median class = 55–65.

Here l = 55, cf = 58, f = 40 and h = 10.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 55 + $\frac{\text{(}\text{70 − 58}\text{)/(}}{\text{40}\text{)}}$ × 10.

Median = 58 years.

No whole-number value satisfies the printed data; formal x = $\frac{\text{(}\text{145}\text{)/(}}{\text{6}\text{)}}$.

The original class marks are 20, 30, 40, 50, 60, 70 and 80.

Σfᵢ = 140 and Σfᵢxᵢ = 7830.

The added people are in class 65–75, whose class mark is 70.

New mean = $\frac{\text{(}\text{7830 + 70x}\text{)/(}}{\text{140 + x}\text{)}}$ = 58.

7830 + 70x = 8120 + 58x.

12x = 290, so x = $\frac{\text{(}\text{145}\text{)/(}}{\text{6}\text{)}}$ ≈ 24.17.

A number of people must be a whole number, so the printed condition has no valid integer solution.

Source note: the 2022 Standard 30/4/1 scheme, PDF and printed p. 11, says an attempted answer “will be awarded full credit of 2 marks”. The fraction above is shown to explain the source issue, not as a valid count of people.

Median = $\frac{\text{(}\text{970}\text{)/(}}{\text{7}\text{)}}$ ≈ 138.57 km.

n = 50; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 25.

The first cumulative frequency greater than 25 is 26.

Median class = 120–140.

Here l = 120, cf = 12, f = 14 and h = 20.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 120 + $\frac{\text{(}\text{25 − 12}\text{)/(}}{\text{14}\text{)}}$ × 20.

Median = $\frac{\text{(}\text{970}\text{)/(}}{\text{7}\text{)}}$ ≈ 138.57 km.

Mean = 145.2 km.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 50.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{7260}\text{)/(}}{\text{50}\text{)}}$ = 145.2 km.

Median = 37.5 years.

Ordinary frequencies are successive differences: 3, 10 − 3 = 7, 22 − 10 = 12, 40 − 22 = 18, 54 − 40 = 14 and 71 − 54 = 17. Since these are non-negative ages and the first cumulative group is ‘less than 10’, the first class is taken as 0–10.

n = 71; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 35.5.

The first cumulative frequency greater than 35.5 is 40.

Median class = 30–40.

Here l = 30, cf = 22, f = 18 and h = 10.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 30 + $\frac{\text{(}\text{35.5 − 22}\text{)/(}}{\text{18}\text{)}}$ × 10.

Median = 37.5 years.

Mean = $\frac{\text{(}\text{2615}\text{)/(}}{\text{71}\text{)}}$ ≈ 36.83 years.

Ordinary frequencies are successive differences: 3, 10 − 3 = 7, 22 − 10 = 12, 40 − 22 = 18, 54 − 40 = 14 and 71 − 54 = 17. Since these are non-negative ages and the first cumulative group is ‘less than 10’, the first class is taken as 0–10.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 71.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{2615}\text{)/(}}{\text{71}\text{)}}$ = $\frac{\text{(}\text{2615}\text{)/(}}{\text{71}\text{)}}$ ≈ 36.83 years.

Median class = 7–10.

n = 50; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 25.

The first cumulative frequency greater than 25 is 27.

Median class = 7–10.

Modal class = 13–16 plants.

The greatest frequency is 12 houses, in the class 13–16 plants.

Median = $\frac{\text{(}\text{28}\text{)/(}}{\text{3}\text{)}}$ ≈ 9.33 plants.

n = 50; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 25.

The first cumulative frequency greater than 25 is 27.

Median class = 7–10.

Here l = 7, cf = 18, f = 9 and h = 3.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 7 + $\frac{\text{(}\text{25 − 18}\text{)/(}}{\text{9}\text{)}}$ × 3.

Median = $\frac{\text{(}\text{28}\text{)/(}}{\text{3}\text{)}}$ ≈ 9.33 plants.

Mode = $\frac{\text{(}\text{184}\text{)/(}}{\text{13}\text{)}}$ ≈ 14.15 plants.

The greatest frequency is 12, so the modal class is 13–16.

Here l = 13, h = 3, f₁ = 12, f₀ = 7 and f₂ = 4.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 13 + $\frac{\text{(}\text{12 − 7}\text{)/(}}{\text{2 × 12 − 7 − 4}\text{)}}$ × 3.

Mode = $\frac{\text{(}\text{184}\text{)/(}}{\text{13}\text{)}}$ ≈ 14.15 plants.

Upper limit = ₹15.

The greatest frequency is 22, so the modal class is 10–15.

Its upper limit is 15, not its frequency 22.

Median class = 10–15.

n = 80; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 40.

The first cumulative frequency greater than 40 is 51.

Median class = 10–15.

Mode = 13 rupees.

The greatest frequency is 22, so the modal class is 10–15.

Here l = 10, h = 5, f₁ = 22, f₀ = 16 and f₂ = 18.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 10 + $\frac{\text{(}\text{22 − 16}\text{)/(}}{\text{2 × 22 − 16 − 18}\text{)}}$ × 5.

Mode = 13 rupees.

Median = 12.5 rupees.

n = 80; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 40.

The first cumulative frequency greater than 40 is 51.

Median class = 10–15.

Here l = 10, cf = 29, f = 22 and h = 5.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 10 + $\frac{\text{(}\text{40 − 29}\text{)/(}}{\text{22}\text{)}}$ × 5.

Median = 12.5 rupees.

Age group = 45–55 years.

The greatest frequency is 23.

It belongs to the class 45–55 years.

Modal age = 47 years; mean age = 45.25 years.

For mode, l = 45, h = 10, f₁ = 23, f₀ = 21 and f₂ = 15.

Mode = 45 + $\frac{\text{(}\text{23 − 21}\text{)/(}}{\text{2 × 23 − 21 − 15}\text{)}}$ × 10 = 47 years.

The class marks are 20, 30, 40, 50, 60 and 70.

Σfᵢxᵢ = 120 + 330 + 840 + 1150 + 900 + 280 = 3620.

Σfᵢ = 80.

Mean = $\frac{\text{(}\text{3620}\text{)/(}}{\text{80}\text{)}}$ = 45.25 years.

Mode = $\frac{\text{(}\text{625}\text{)/(}}{\text{17}\text{)}}$ ≈ 36.76 years.

The greatest frequency is 33, so the modal class is 35–40.

Here l = 35, h = 5, f₁ = 33, f₀ = 21 and f₂ = 11.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 35 + $\frac{\text{(}\text{33 − 21}\text{)/(}}{\text{2 × 33 − 21 − 11}\text{)}}$ × 5.

Mode = $\frac{\text{(}\text{625}\text{)/(}}{\text{17}\text{)}}$ ≈ 36.76 years.

Median = $\frac{\text{(}\text{1180}\text{)/(}}{\text{33}\text{)}}$ ≈ 35.76 years.

n = 100; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 50.

The first cumulative frequency greater than 50 is 78.

Median class = 35–40.

Here l = 35, cf = 45, f = 33 and h = 5.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 35 + $\frac{\text{(}\text{50 − 45}\text{)/(}}{\text{33}\text{)}}$ × 5.

Median = $\frac{\text{(}\text{1180}\text{)/(}}{\text{33}\text{)}}$ ≈ 35.76 years.

Median class = 46–48 kg; median = 46.5 kg.

n = 35; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 17.5.

The first cumulative frequency greater than 17.5 is 28.

Median class = 46–48.

Here l = 46, cf = 14, f = 14 and h = 2.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 46 + $\frac{\text{(}\text{17.5 − 14}\text{)/(}}{\text{14}\text{)}}$ × 2.

Median = 46.5 kg.

Modal class = 46–48 kg; mode = 46.9 kg.

The greatest frequency is 14, so the modal class is 46–48.

Here l = 46, h = 2, f₁ = 14, f₀ = 5 and f₂ = 3.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 46 + $\frac{\text{(}\text{14 − 5}\text{)/(}}{\text{2 × 14 − 5 − 3}\text{)}}$ × 2.

Mode = 46.9 kg.

Median class = 100–110.

n = 40; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 20.

The first cumulative frequency greater than 20 is 29.

Median class = 100–110.

Number of leaves = 23.

10 cm = 100 mm.

Use the classes 100–110, 110–120, 120–130 and 130–140 mm.

Number = 12 + 5 + 4 + 2 = 23.

Median = 102.5 mm.

n = 40; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 20.

The first cumulative frequency greater than 20 is 29.

Median class = 100–110.

Here l = 100, cf = 17, f = 12 and h = 10.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 100 + $\frac{\text{(}\text{20 − 17}\text{)/(}}{\text{12}\text{)}}$ × 10.

Median = 102.5 mm.

Modal class = 100–110 mm; mode = 103 mm.

The greatest frequency is 12, so the modal class is 100–110.

Here l = 100, h = 10, f₁ = 12, f₀ = 9 and f₂ = 5.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 100 + $\frac{\text{(}\text{12 − 9}\text{)/(}}{\text{2 × 12 − 9 − 5}\text{)}}$ × 10.

Mode = 103 mm.

Number of women = 15.

Add the frequencies for 68–71, 71–74 and 74–77.

Number = 4 + 3 + 8 = 15.

Median class = 74–77.

n = 30; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 15.

The first cumulative frequency greater than 15 is 17.

Median class = 74–77.

Mode = 76.5 beats per minute.

The greatest frequency is 8, so the modal class is 74–77.

Here l = 74, h = 3, f₁ = 8, f₀ = 3 and f₂ = 7.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 74 + $\frac{\text{(}\text{8 − 3}\text{)/(}}{\text{2 × 8 − 3 − 7}\text{)}}$ × 3.

Mode = 76.5 beats per minute.

Median = 76.25 beats per minute.

n = 30; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 15.

The first cumulative frequency greater than 15 is 17.

Median class = 74–77.

Here l = 74, cf = 9, f = 8 and h = 3.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 74 + $\frac{\text{(}\text{15 − 9}\text{)/(}}{\text{8}\text{)}}$ × 3.

Median = 76.25 beats per minute.

Upper limit = 45 years.

Hospital I’s greatest frequency is 23, in class 35–45.

The upper class limit is 45 years.

Lower limit = 25 years.

Hospital II’s greatest frequency is 32, in class 25–35.

The lower class limit is 25 years.

Mode = $\frac{\text{(}\text{405}\text{)/(}}{\text{11}\text{)}}$ ≈ 36.82 years.

The greatest frequency is 23, so the modal class is 35–45.

Here l = 35, h = 10, f₁ = 23, f₀ = 21 and f₂ = 14.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 35 + $\frac{\text{(}\text{23 − 21}\text{)/(}}{\text{2 × 23 − 21 − 14}\text{)}}$ × 10.

Mode = $\frac{\text{(}\text{405}\text{)/(}}{\text{11}\text{)}}$ ≈ 36.82 years.

Mode = $\frac{\text{(}\text{555}\text{)/(}}{\text{19}\text{)}}$ ≈ 29.21 years.

The greatest frequency is 32, so the modal class is 25–35.

Here l = 25, h = 10, f₁ = 32, f₀ = 16 and f₂ = 10.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 25 + $\frac{\text{(}\text{32 − 16}\text{)/(}}{\text{2 × 32 − 16 − 10}\text{)}}$ × 10.

Mode = $\frac{\text{(}\text{555}\text{)/(}}{\text{19}\text{)}}$ ≈ 29.21 years.

Lower limit = 35 students per teacher.

Cumulative frequencies are 5, 20, 45, 75, 90 and 100.

n/2 = 50, so the median class is 35–40.

Its lower limit is 35.

Upper limit = 40 students per teacher.

The greatest frequency is 30, in class 35–40.

Its upper limit is 40.

Median = $\frac{\text{(}\text{215}\text{)/(}}{\text{6}\text{)}}$ ≈ 35.83 students per teacher.

n = 100; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 50.

The first cumulative frequency greater than 50 is 75.

Median class = 35–40.

Here l = 35, cf = 45, f = 30 and h = 5.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 35 + $\frac{\text{(}\text{50 − 45}\text{)/(}}{\text{30}\text{)}}$ × 5.

Median = $\frac{\text{(}\text{215}\text{)/(}}{\text{6}\text{)}}$ ≈ 35.83 students per teacher.

Mode = 36.25 students per teacher.

The greatest frequency is 30, so the modal class is 35–40.

Here l = 35, h = 5, f₁ = 30, f₀ = 25 and f₂ = 15.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 35 + $\frac{\text{(}\text{30 − 25}\text{)/(}}{\text{2 × 30 − 25 − 15}\text{)}}$ × 5.

Mode = 36.25 students per teacher.

Modal class = 30–35 students per teacher.

The greatest frequency is 10 schools, in the class 30–35.

Median class = 25–30.

n = 35; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 17.5.

The first cumulative frequency greater than 17.5 is 20.

Median class = 25–30.

Mode = 30.625 students per teacher.

The greatest frequency is 10, so the modal class is 30–35.

Here l = 30, h = 5, f₁ = 10, f₀ = 9 and f₂ = 3.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 30 + $\frac{\text{(}\text{10 − 9}\text{)/(}}{\text{2 × 10 − 9 − 3}\text{)}}$ × 5.

Mode = 30.625 students per teacher.

Median = $\frac{\text{(}\text{515}\text{)/(}}{\text{18}\text{)}}$ ≈ 28.61 students per teacher.

n = 35; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 17.5.

The first cumulative frequency greater than 17.5 is 20.

Median class = 25–30.

Here l = 25, cf = 11, f = 9 and h = 5.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 25 + $\frac{\text{(}\text{17.5 − 11}\text{)/(}}{\text{9}\text{)}}$ × 5.

Median = $\frac{\text{(}\text{515}\text{)/(}}{\text{18}\text{)}}$ ≈ 28.61 students per teacher.

Official CBSE answer: 19.5 years, using continuous class boundaries.

The greatest frequency is 132, so the printed modal class is 20–24.

The neighbouring inclusive classes 15–19 and 20–24 have a gap of 1.

Half the gap is 0.5; the continuous modal class is 19.5–24.5.

Its lower boundary is 19.5 years.

Source note: all three 2024 Standard schemes, 30/3/1, 30/3/2 and 30/3/3, give 19.5 on PDF and printed page 19. They interpret the requested lower limit after conversion to continuous classes. The printed class limit remains 20; distinguish it from the continuous boundary.

Median class = 19.5–24.5 years in continuous form (printed class: 20–24).

n = 62 + 132 + 96 + 37 + 13 + 11 + 10 + 4 = 365.

$\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 182.5.

Cumulative frequencies begin 62, 194, 290, … .

The first cumulative frequency greater than 182.5 is 194, in the printed class 20–24.

After conversion, the median class is 19.5–24.5, as shown in the official schemes on PDF and printed page 20.

Number of participants = 361.

The last class, 50–54, contains 4 participants.

Number younger than 50 = 365 − 4 = 361.

Mode ≈ 3 × Median − 2 × Mean.

The empirical relationship is commonly written Mode = 3 Median − 2 Mean when used in a question.

It is an approximate relationship, not an exact identity for every distribution.

Median class = 30–45.

n = 48; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 24.

The first cumulative frequency greater than 24 is 27.

Median class = 30–45.

Probability = $\frac{\text{(}\text{37}\text{)/(}}{\text{75}\text{)}}$.

The even numbers are 2, 4, 6, …, 74: there are 37.

There are 75 possible balls on the first draw.

P(even number) = $\frac{\text{(}\text{37}\text{)/(}}{\text{75}\text{)}}$.

Median = 40.5.

n = 48; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 24.

The first cumulative frequency greater than 24 is 27.

Median class = 30–45.

Here l = 30, cf = 17, f = 10 and h = 15.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 30 + $\frac{\text{(}\text{24 − 17}\text{)/(}}{\text{10}\text{)}}$ × 15.

Median = 40.5.

Mode = 51.

The greatest frequency is 12, so the modal class is 45–60.

Here l = 45, h = 15, f₁ = 12, f₀ = 10 and f₂ = 9.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 45 + $\frac{\text{(}\text{12 − 10}\text{)/(}}{\text{2 × 12 − 10 − 9}\text{)}}$ × 15.

Mode = 51.

Median class = 40–60.

n = 40; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 20.

The first cumulative frequency greater than 20 is 31.

Median class = 40–60.

Mean = 43 seconds.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 40.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{1720}\text{)/(}}{\text{40}\text{)}}$ = 43 seconds.

Mode = 46 seconds.

The greatest frequency is 13, so the modal class is 40–60.

Here l = 40, h = 20, f₁ = 13, f₀ = 10 and f₂ = 6.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 40 + $\frac{\text{(}\text{13 − 10}\text{)/(}}{\text{2 × 13 − 10 − 6}\text{)}}$ × 20.

Mode = 46 seconds.

Number of students = 31.

Add the first three class frequencies.

Number = 8 + 10 + 13 = 31.

Most common performance range = 30–40 marks.

The greatest frequency is 15 students in class 30–40.

This is the modal class, not the numerical mode.

Median class = 30–40.

n = 50; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 25.

The first cumulative frequency greater than 25 is 36.

Median class = 30–40.

Mean = 31.2 marks.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 50.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{1560}\text{)/(}}{\text{50}\text{)}}$ = 31.2 marks.

Mode = 37.5 marks.

The greatest frequency is 15, so the modal class is 30–40.

Here l = 30, h = 10, f₁ = 15, f₀ = 12 and f₂ = 14.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 30 + $\frac{\text{(}\text{15 − 12}\text{)/(}}{\text{2 × 15 − 12 − 14}\text{)}}$ × 10.

Mode = 37.5 marks.

Modal class = 600–800 mm.

The greatest frequency is 7, corresponding to 600–800 mm.

Median = $\frac{\text{(}\text{5400}\text{)/(}}{\text{7}\text{)}}$ ≈ 771.43 mm.

n = 24; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 12.

The first cumulative frequency greater than 12 is 13.

Median class = 600–800.

Here l = 600, cf = 6, f = 7 and h = 200.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 600 + $\frac{\text{(}\text{12 − 6}\text{)/(}}{\text{7}\text{)}}$ × 200.

Median = $\frac{\text{(}\text{5400}\text{)/(}}{\text{7}\text{)}}$ ≈ 771.43 mm.

Mean = 850 mm.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 24.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{20400}\text{)/(}}{\text{24}\text{)}}$ = 850 mm.

Good-rainfall sub-divisions = 7.

At least 1000 mm includes the classes starting at 1000 mm and above.

Number = 2 + 3 + 1 + 1 = 7.

Modal class = 600–800 mm.

The greatest frequency is 7, in class 600–800 mm.

Median = $\frac{\text{(}\text{5200}\text{)/(}}{\text{7}\text{)}}$ ≈ 742.86 mm.

n = 24; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 12.

The first cumulative frequency greater than 12 is 14.

Median class = 600–800.

Here l = 600, cf = 7, f = 7 and h = 200.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 600 + $\frac{\text{(}\text{12 − 7}\text{)/(}}{\text{7}\text{)}}$ × 200.

Median = $\frac{\text{(}\text{5200}\text{)/(}}{\text{7}\text{)}}$ ≈ 742.86 mm.

Mean = 775 mm.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 24.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{18600}\text{)/(}}{\text{24}\text{)}}$ = 775 mm.

Good-rainfall sub-divisions = 10.

Use the classes 800–1000, 1000–1200 and 1200–1400 mm.

Number = 4 + 3 + 3 = 10.

Continuous classes: 0.5–100.5, 100.5–200.5, 200.5–300.5, 300.5–400.5, 400.5–500.5.

The gap between successive class limits is 101 − 100 = 1.

Half-gap correction = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ = 0.5.

Subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Frequencies remain unchanged.

Air quality: Poor.

The greatest frequency is 12 days in the original interval 201–300.

The printed reference chart labels 201–300 as Poor.

Mode = $\frac{\text{(}\text{5011}\text{)/(}}{\text{22}\text{)}}$ ≈ 227.77.

The greatest frequency is 12, so the modal class is 200.5–300.5.

Here l = 200.5, h = 100, f₁ = 12, f₀ = 9 and f₂ = 4.

Mode = l + $\frac{\text{(}\text{f₁ − f₀}\text{)/(}}{\text{2f₁ − f₀ − f₂}\text{)}}$ × h.

Mode = 200.5 + $\frac{\text{(}\text{12 − 9}\text{)/(}}{\text{2 × 12 − 9 − 4}\text{)}}$ × 100.

Mode = $\frac{\text{(}\text{5011}\text{)/(}}{\text{22}\text{)}}$ ≈ 227.77.

Median = 225.5.

n = 30; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 15.

The first cumulative frequency greater than 15 is 24.

Median class = 200.5–300.5.

Here l = 200.5, cf = 12, f = 12 and h = 100.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 200.5 + $\frac{\text{(}\text{15 − 12}\text{)/(}}{\text{12}\text{)}}$ × 100.

Median = 225.5.

Lower limit = 0.08 ppm.

Cumulative frequencies are 4, 12, 22, 24, 28 and 30.

n/2 = 15. The first cumulative frequency greater than 15 is 22.

The median class is 0.08–0.12, whose lower limit is 0.08.

Modal class = 0.08–0.12 ppm.

The greatest frequency is 10, in class 0.08–0.12 ppm.

Mean = 0.1 ppm.

Class mark xᵢ = $\frac{\text{(}\text{lower limit + upper limit}\text{)/(}}{\text{2}\text{)}}$.

Total frequency n = Σfᵢ = 30.

Mean x̄ = $\frac{\text{(}\text{Σfᵢxᵢ}\text{)/(}}{\text{Σfᵢ}\text{)}}$.

x̄ = $\frac{\text{(}\text{3}\text{)/(}}{\text{30}\text{)}}$ = 0.1 ppm.

Median = 0.092 ppm.

n = 30; $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$ = 15.

The first cumulative frequency greater than 15 is 22.

Median class = 0.08–0.12.

Here l = 0.08, cf = 12, f = 10 and h = 0.04.

Median = l + $\frac{\text{(}\text{n/2 − cf}\text{)/(}}{\text{f}\text{)}}$ × h.

Median = 0.08 + $\frac{\text{(}\text{15 − 12}\text{)/(}}{\text{10}\text{)}}$ × 0.04.

Median = 0.092 ppm.