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Question

Using the frequency distribution table given below, draw 'less than ogive'. Then from the ogive, find:

interquartile range
Class intervalFrequency10−20220−30430−40640−50850−6010


A

21

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B

22

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C

21.5

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D

22.5

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Solution

The correct option is A

21


We prepare the cumulative frequency table as given below:
Class intervalFrequencyCumulativefrequency1020222030463040612405082050601030hline
We need to plot the points (10, 0), (20, 2), (30, 6), (40, 12), (50, 20) and (60, 30). Plotting these points on a graph paper and joining by a freehand curve, we get the cumulative frequency curve (a 'less than ogive') as given below:

Finding the lower quartile and upper quartile to find the inter-quartile range

i) To find the lower quartile Q1, we locate n+14, i.e., 314 =7.5 on y-axis and proceed horizontally to meet the ogive. From this point, draw a perpendicular to meet the x-axis.
From figure Q1=32.5

ii) To find the upper quartile Q3, we proceed similarly as in (i) taking
3(n+1)4=(3×7.754)=23.25 on y-axis
Then, Q3=53.5.

iii) Interquartile range Q3Q1= 53.5 - 32.5 = 21


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