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Worked Examples · Example 15
Q.

Prepare the histogram for following data:

Marks0-55-1010-2020-4040-5050-80
No of students791412815

(Class intervals are unequal, so adjust the frequencies before constructing the histogram: Adjusted frequency of a class = (Minimum class size) / (Class size) × Frequency.)

Chandigarh CbseNCERTSubjective· 3mImportance★★★★★est
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With unequal class widths, first compute Adjusted Frequency =Minimum class sizeClass size×Frequency=\dfrac{\text{Minimum class size}}{\text{Class size}}\times \text{Frequency} for every class, then draw the histogram using these adjusted heights.

[!FORMULA] When class widths are UNEQUAL, the histogram bar height is not the raw frequency but the Adjusted Frequency:

Adjusted Frequency=Minimum class sizeClass size of that interval×Frequency of that interval\text{Adjusted Frequency} = \dfrac{\text{Minimum class size}}{\text{Class size of that interval}}\times \text{Frequency of that interval}

This prevents wider classes from visually appearing to have disproportionately "taller" bars just because they span more marks.

  1. List the class intervals, widths and raw frequencies:
Marks0-55-1010-2020-4040-5050-80
Class size5510201030
No. of students (freq.)791412815
  1. Find the minimum class size: comparing 5,5,10,20,10,305, 5, 10, 20, 10, 30, the minimum is 55 (from 0-5 or 5-10).
  2. Apply the adjustment formula to each class, Adjusted Freq.=5class size×frequency\text{Adjusted Freq.} = \dfrac{5}{\text{class size}}\times \text{frequency}:
    • 00-55: 55×7=7\dfrac{5}{5}\times 7 = 7
    • 55-1010: 55×9=9\dfrac{5}{5}\times 9 = 9
    • 1010-2020: 510×14=7\dfrac{5}{10}\times 14 = 7
    • 2020-4040: 520×12=3\dfrac{5}{20}\times 12 = 3
    • 4040-5050: 510×8=4\dfrac{5}{10}\times 8 = 4
    • 5050-8080: 530×15=2.5\dfrac{5}{30}\times 15 = 2.5 …

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