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Worked Examples · Example 16
Q.

Consider the marks, out of 100, obtained by 70 students of a class in a test, given in table below:

Class Interval10-2020-3030-4040-5050-6060-70
Frequency5121520144

Draw a frequency polygon corresponding to this frequency distribution table (without drawing the histogram, using class marks Class Mark=Upper Limit+Lower Limit2\text{Class Mark} = \dfrac{\text{Upper Limit} + \text{Lower Limit}}{2}).

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
49% · 28/57 Questions
✓ Free question

Compute the class mark of every class, add two zero-frequency classes at the ends, then plot (class mark, frequency) and join the points by straight-line segments — no histogram is drawn.

[!FORMULA] Class Mark (x)=Upper Limit+Lower Limit2\text{Class Mark } (x) = \dfrac{\text{Upper Limit} + \text{Lower Limit}}{2}

where Upper Limit and Lower Limit are the boundaries of each class interval. The frequency polygon plots (x,f)(x, f) for each class.

  1. List the classes and frequencies given:

    Class IntervalFrequency (f)(f)
    10–205
    20–3012
    30–4015
    40–5020
    50–6014
    60–704
  2. Compute the class mark of each class using x=Upper Limit+Lower Limit2x = \dfrac{\text{Upper Limit}+\text{Lower Limit}}{2}:

    x1=10+202=15, x2=20+302=25, x3=30+402=35,x_1=\dfrac{10+20}{2}=15,\ x_2=\dfrac{20+30}{2}=25,\ x_3=\dfrac{30+40}{2}=35,

    x4=40+502=45, x5=50+602=55, x6=60+702=65x_4=\dfrac{40+50}{2}=45,\ x_5=\dfrac{50+60}{2}=55,\ x_6=\dfrac{60+70}{2}=65

  3. Add an imaginary class of zero frequency on each side so the polygon touches the x-axis: class 00–1010 (mark 55, frequency 00) before the data, and class 7070–8080 (mark 7575, frequency 00) after the data.

  4. Tabulate every point (x,f)(x,f) to be plotted:

    Class Mark (x)(x)515253545556575
    Frequency (f)(f)051215201440
  5. Plot the points on graph paper (class marks on the x-axis, frequency on the y-axis) and join consecutive points by straight-line segments. The closed figure obtained is the frequency polygon; it rises to a peak at (45,20)(45,20) (the class 4040–5050, the highest frequency) and returns to zero at both ends.

✓Final answer

Frequency polygon through (5,0),(15,5),(25,12),(35,15),(45,20),(55,14),(65,4),(75,0)(5,0),(15,5),(25,12),(35,15),(45,20),(55,14),(65,4),(75,0), peaking at class mark 4545 (frequency 2020).

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