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Question

Calculate the median and mode from the following data: Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Number of Students 2 18 30 45 35 20 6 3

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Solution

Marks No. of Workers (f) Cumulative Frequency (c.f.) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 2 18 30 = f0 45 = f1 35 = f2 20 6 3 2 20 50 95 130 150 156 159 N = ∑f = 159 $\overline{)\mathbf{Median}}$ Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{159}{2}\right)}^{th}$ item, which is 79.5th item. This corresponds to the class interval of (30 − 40), so this is the median class. $\mathrm{Median}={l}_{1}+\frac{\frac{N}{2}-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}\mathrm{so},\mathrm{Median}=30+\frac{\frac{159}{2}-50}{45}×i\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{Median}=30+\frac{29.5}{45}×10\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{Median}=30+6.55\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Median}=36.55$ $\overline{)\mathbf{Mode}}$ By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 45. $\mathrm{Mode}\left(Z\right)={l}_{1}+\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}×i\phantom{\rule{0ex}{0ex}}\mathrm{or},Z=30+\frac{45-30}{2\left(45\right)-30-35}×10\phantom{\rule{0ex}{0ex}}\mathrm{or},Z=30+\frac{15}{90-65}×10\phantom{\rule{0ex}{0ex}}\mathrm{or},Z=30+\frac{150}{25}=30+6=36\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{mode}\left(Z\right)\mathrm{is}36$

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