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Question

# Find the mean, mode and median of the following frequency distribution. [CBSE 2010] Class 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency 4 4 7 10 12 8 5

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Solution

## To find the mean let us put the data in the table given below: Class Frequency (fi) Class mark (xi) fixi 0−10 4 5 20 10−20 4 15 60 20−30 7 25 175 30−40 10 35 350 40−50 12 45 540 50−60 8 55 440 60−70 5 65 325 Total ∑fi = 50 ∑fixi = 1910 $\mathrm{Mean}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}=\frac{1910}{50}\phantom{\rule{0ex}{0ex}}=38.2$ Thus, mean of the given data is 38.2. Now, to find the median let us put the data in the table given below: Class Frequency (fi) Cumulative frequency (cf) 0−10 4 4 10−20 4 8 20−30 7 15 30−40 10 25 40−50 12 37 50−60 8 45 60−70 5 50 Total N = ∑fi = 50 Now, N = 50 $⇒\frac{N}{2}=25$. The cumulative frequency just greater than 25 is 37, and the corresponding class is 40−50. Thus, the median class is 40−50. ∴ l = 40, h = 10, N = 50, f = 12 and cf = 25. Now, $\mathrm{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)×h\phantom{\rule{0ex}{0ex}}=40+\left(\frac{25-25}{12}\right)×10\phantom{\rule{0ex}{0ex}}=40$ Thus, the median is 40. We know that, Mode = 3(median) − 2(mean) = 3 × 40 − 2 × 38.2 = 120 − 76.4 = 43.6 Hence, Mean = 38.2, Median = 40 and Mode = 43.6

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