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Worked Examples · Example 17
Q.

Construct frequency polygon for the following set of data:

Class Interval0-1010-2020-3030-4040-5050-6060-70
Frequency457101284

(Construct it by first drawing the histogram, then joining the class marks located on the top horizontal side of each rectangle, adding two imaginary zero-frequency intervals before and after the data.)

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First sketch the histogram for the distribution, then locate the midpoint of the top of every bar (its class mark) and join these midpoints — plus two zero-frequency midpoints at the ends — with straight lines to obtain the frequency polygon.

[!FORMULA] Class Mark (x)=Upper Limit+Lower Limit2\text{Class Mark }(x)=\dfrac{\text{Upper Limit}+\text{Lower Limit}}{2}; the histogram bar over each class has height equal to its frequency ff, and the midpoint of the top edge of that bar is the point (x,f)(x,f).

  1. Tabulate the given data:

    Class IntervalFrequency (f)(f)
    0–104
    10–205
    20–307
    30–4010
    40–5012
    50–608
    60–704
  2. Draw the histogram: on graph paper, draw seven adjoining rectangles, one per class interval, each with height equal to its frequency (the bar over 00–1010 has height 44, over 1010–2020 height 55, and so on up to 6060–7070 with height 44).

  3. Mark the midpoint of the top edge of every bar. Its x-coordinate is the class mark x=Upper Limit+Lower Limit2x=\dfrac{\text{Upper Limit}+\text{Lower Limit}}{2}:

    x1=0+102=5, x2=10+202=15, x3=20+302=25, x4=30+402=35,x_1=\dfrac{0+10}{2}=5,\ x_2=\dfrac{10+20}{2}=15,\ x_3=\dfrac{20+30}{2}=25,\ x_4=\dfrac{30+40}{2}=35,

    x5=40+502=45, x6=50+602=55, x7=60+702=65x_5=\dfrac{40+50}{2}=45,\ x_6=\dfrac{50+60}{2}=55,\ x_7=\dfrac{60+70}{2}=65 …

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