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Worked Examples · Example 30
Q.

Calculate the Quartile Deviation from following frequency distribution:

Class Interval30-3435-3940-4445-4950-5455-5960-6465-6970-7475-7980-84
Frequency126791187531
CBSENCERTSubjective· 3mImportance★★★★★est
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Convert the class intervals to continuous boundaries, locate the Q1Q_1-class and Q3Q_3-class from the cumulative frequency, interpolate each quartile, and take half their difference.

[!FORMULA] Qk=l+(kN4−cf)f×h,QD=Q3−Q12Q_k=l+\dfrac{\left(\dfrac{kN}{4}-cf\right)}{f}\times h,\qquad QD=\dfrac{Q_3-Q_1}{2}

where ll = lower boundary of the quartile class, N=∑fN=\sum f, cfcf = cumulative frequency before the quartile class, ff = frequency of the quartile class, hh = class width, k=1k=1 for Q1Q_1 and k=3k=3 for Q3Q_3.

  1. Convert the given (inclusive) classes to continuous class boundaries (subtract 0.50.5 from each lower limit, add 0.50.5 to each upper limit), and tabulate cumulative frequency:

    Class Boundariesffcfcf
    29.5–34.511
    34.5–39.523
    39.5–44.569
    44.5–49.5716
    49.5–54.5925
    54.5–59.51136
    59.5–64.5844
    64.5–69.5751
    69.5–74.5556
    74.5–79.5359
    79.5–84.5160
  2. Compute N=∑f=1+2+6+7+9+11+8+7+5+3+1=60N=\sum f=1+2+6+7+9+11+8+7+5+3+1=60, class width h=5h=5.

  3. Find the Q1Q_1 class: N4=604=15\dfrac{N}{4}=\dfrac{60}{4}=15. The first cf≥15cf\ge15 is 1616, in class 44.544.5–49.549.5, with cfcf(before)=9=9, f=7f=7, l=44.5l=44.5. …

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