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Difficulty: Very hardData Representation and Charts

The grouped frequency table below shows the distribution of marks obtained by candidates in a Mathematics examination:

Class IntervalFrequency
10 – 1915
20 – 2925
30 – 39kk
40 – 4920
50 ��� 5910

When this data is represented on a pie chart, the sector corresponding to the score range 30 – 39 has a central angle of 108∘108^\circ. Based on a cumulative frequency curve (ogive) constructed for this distribution, what is the score corresponding to the 75th percentile (Q3Q_3) of the candidates?

  1. 42.0Answer
  2. B
    42.5
  3. C
    41.5
  4. D
    44.5

Answer

42.0
The value 42.0 is obtained by first determining the missing frequency k=30k = 30 using the ratio 108∘360∘=0.3\frac{108^\circ}{360^\circ} = 0.3, establishing N=100N = 100. The 75th percentile position is at 75, which falls in the 40–4940 – 49 class. Interpolating from the lower boundary of 39.5 yields 39.5+(75−7020)×10=42.039.5 + \left(\frac{75 - 70}{20}\right) \times 10 = 42.0.

Step-by-Step Solution

1
Determine the unknown frequency kk using the pie chart sector angle.
kTotal Frequency=108∘360∘=0.3 \frac{k}{\text{Total Frequency}} = \frac{108^\circ}{360^\circ} = 0.3
Total frequency N=15+25+k+20+10=70+kN = 15 + 25 + k + 20 + 10 = 70 + k.
k70+k=0.3  ⟹  k=21+0.3k  ⟹  0.7k=21  ⟹  k=30 \frac{k}{70 + k} = 0.3 \implies k = 21 + 0.3k \implies 0.7k = 21 \implies k = 30
The sector angle in a pie chart is proportional to the category frequency relative to the total frequency across 360∘360^\circ.
2
Calculate cumulative frequencies and locate the 75th percentile position.
Total frequency N=100N = 100.
Cumulative frequencies (cfcf):
- 10–1910 – 19 (boundary 9.5–19.59.5 – 19.5): cf=15cf = 15
- 20–2920 – 29 (boundary 19.5–29.519.5 – 29.5): cf=40cf = 40
- 30–3930 – 39 (boundary 29.5–39.529.5 – 39.5): cf=70cf = 70
- 40–4940 – 49 (boundary 39.5–49.539.5 – 49.5): cf=90cf = 90
- 50–5950 – 59 (boundary 49.5–59.549.5 – 59.5): cf=100cf = 100

75th percentile position =0.75×100=75th candidate= 0.75 \times 100 = 75\text{th candidate}.
The 75th percentile corresponds to the value below which 75% of the total observations lie.
3
Apply linear interpolation on the percentile class interval 40–4940 – 49.
The 75th score lies in class interval 40–4940 – 49 (boundaries 39.5–49.539.5 – 49.5).
- Lower class boundary L=39.5L = 39.5
- Cumulative frequency prior to class cfb=70cf_b = 70
- Frequency of percentile class f=20f = 20
- Class width c=10c = 10

Q3=L+(0.75N−cfbf)×c=39.5+(75−7020)×10=39.5+2.5=42.0 Q_3 = L + \left(\frac{0.75N - cf_b}{f}\right) \times c = 39.5 + \left(\frac{75 - 70}{20}\right) \times 10 = 39.5 + 2.5 = 42.0
Linear interpolation along an ogive requires using exact class boundaries to compute specific percentile values.

Key Concept

Pie chart sector angles and linear interpolation on cumulative frequency distributions
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Topics

Data Representation and Charts

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