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Difficulty: Very hardData Representation and Charts

The grouped frequency distribution table below shows the marks scored by 100100 candidates in a Mathematics screening test:

Class IntervalFrequency
10−1910 - 1915
20−2920 - 2925
30−3930 - 3940
40−4940 - 4920

Match each data representation component on the left with its correct calculated numerical value on the right.

  • Sector angle representing the modal class in a pie chart144∘144^\circ
  • Frequency density of the modal class for a histogram4.04.0
  • Upper class boundary of the class interval immediately preceding the modal class29.529.5
  • Cumulative frequency corresponding to the upper boundary of the median class on an ogive8080

Answer

The sector angle matches 144∘144^\circ, frequency density matches 4.04.0, preceding upper class boundary matches 29.529.5, and median class cumulative frequency matches 8080.
The items match based on direct statistical computations: the modal sector angle is 144∘144^\circ, frequency density is 4.04.0, upper class boundary of the preceding interval is 29.529.5, and cumulative frequency at the upper boundary of the median class is 8080.

Step-by-Step Solution

1
Determine the modal class and total frequency NN.
The highest frequency is 4040, so the modal class is 30−3930 - 39. Total frequency N=15+25+40+20=100N = 15 + 25 + 40 + 20 = 100.
Modal class identification is essential for pie chart sector, frequency density, and class boundary calculations.
2
Calculate the sector angle for the modal class in a pie chart.
Sector Angle =40100×360∘=144∘= \frac{40}{100} \times 360^\circ = 144^\circ.
The sector angle represents the class frequency as a fraction of total frequency multiplied by 360∘360^\circ.
3
Calculate the frequency density of the modal class.
Class boundary range for 30−3930 - 39 is 29.5−39.529.5 - 39.5, so width =10= 10. Frequency density =4010=4.0= \frac{40}{10} = 4.0.
Frequency density is defined as class frequency divided by class interval width.
4
Determine the preceding upper class boundary and the cumulative frequency for the median class.
Preceding interval is 20−2920 - 29, so its upper boundary is 29.529.5. Median is at position 5050 (in interval 30−3930 - 39). Cumulative frequency up to boundary 39.539.5 is 15+25+40=8015 + 25 + 40 = 80.
Class boundaries are midpoints between adjacent class limits, and cumulative frequency sums all preceding frequencies up to the upper boundary.

Key Concept

Interpretation and calculation of pie chart sector angles, histogram frequency densities, real class boundaries, and cumulative frequencies from grouped data.
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Topics

Data Representation and Charts

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