Statistics and Probability · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.
1. Which formula is used to find the median of grouped data?
- Al+fn/2−cf×h
- Bl+cff−n/2×h
- Ca+∑f∑f⋅d
- Dl+2f1−f0−f2f1−f0×h
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Answer: A) l+fn/2−cf×h
The median of grouped data is given by Median = l+fn/2−cf×h, where l is the lower boundary of the median class, cf is the cumulative frequency before it, f is its frequency and h is the class size.
2. For the marks distribution
0-10 -> 3, 10-20 -> 5, 20-30 -> 12, 30-40 -> 6, 40-50 -> 4,
compute the median.
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Answer: C) 25.83
n = 30, 2n=15. Median class 20-30, l = 20, cf = 8, f = 12, h = 10. Median = 20+1215−8×10=20+5.833=25.83 (approx).
3. An inclusive class is written as 21-30. For computing the median we must use its true (continuous) lower boundary instead of 21.
What is the true lower boundary of this class?
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Answer: A) 20.5
To make classes continuous, subtract half the gap (21=0.5) from the stated lower limit: 21 - 0.5 = 20.5. This 20.5 is the value of l used in the median formula.
4. The ages (in years) of 150 people attending a health camp, grouped in unequal classes:
0-5 -> 18, 5-15 -> 32, 15-35 -> 45, 35-60 -> 37, 60-90 -> 18.
Which class contains the median, and what class size h goes with it?
- A15-35 with h = 20
- B5-15 with h = 10
- C35-60 with h = 25
- D15-35 with h = 10
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Answer: A) 15-35 with h = 20
Cumulative: 18, 50, 95, 132, 150, so n = 150 and 2n=75. The first cf to reach 75 is 95, so the median lies in 15-35, whose own width is 35−15=20. 5-15 is the row that supplies cf = 50, not the median class. 35-60 is one row too far. The last option keeps the right class but borrows h = 10 from the second class — h must always be the median class's own width, especially when the widths are unequal.
5. A table of monthly incomes has open-ended first and last classes:
below 10 -> 8, 10-20 -> 17, 20-30 -> 25, 30-40 -> 14, 40 and above -> 6.
Can the median be found, and if so what is it?
- AYes, the median is 24 — the open-ended classes are never the median class, so their widths are not needed
- BNo — the widths of the open-ended classes are unknown, so the median cannot be computed
- CYes, the median is 25, the class mark of the median class
- DYes, the median is 14, taking cf = 50
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Answer: A) Yes, the median is 24 — the open-ended classes are never the median class, so their widths are not needed
Cumulative frequencies 8, 25, 50, 64, 70 give n = 70 and 2n=35. The first cf to reach 35 is 50, so the median class is 20-30 — an ordinary closed class. Every ingredient is known: Median = 20+(2535−25)×10=24. Open-ended classes contribute only their frequencies to the cumulative column, so they can never block a median; contrast the MEAN, which does need a class mark for every class and therefore cannot be computed here without assuming widths. 25 is the class mark of 20-30, not the median, and 14 comes from using cf = 50, the median class's own cumulative frequency.