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Question

# Calculate median from the following figures: Class-intervals 10−29 30−39 40−49 50−59 59−60 60−69 Frequencies 12 19 20 21 15 13

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Solution

## Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula. $\mathrm{Value}\mathrm{of}\mathrm{Adjustment}=\frac{\mathrm{Lower}\mathrm{limit}\mathrm{of}\mathrm{one}\mathrm{class}-\mathrm{Upper}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{preceeding}\mathrm{class}}{2}$ Value of lower limit of one class − Value of upper limit of the preceeding class2 The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution. Inclusive Class Interval Exclusive Class Interval Frequency (f) Cumulative Frequency (c.f.) 10 − 19 20 − 29 9.5 − 19.5 19.5 − 29.5 12 19 12 31 (c.f.) 30 − 39 29.5 − 39.5 20 (f) 51 40 − 49 50 − 59 60 − 69 39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 21 15 13 72 87 100 $N={\sum }_{}^{}f=100$ Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item. This corresponds to the class interval of (29.5 − 39.5), so this is the median class. $\mathrm{Median}={l}_{1}+\frac{\frac{N}{2}-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}\mathrm{so},\mathrm{Median}=29.5+\frac{\frac{100}{2}-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{Median}=29.5+\frac{50-31}{20}×10\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{Median}=29.5+9.5=39\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Median}=39$

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