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# The mean of the following data is 42, find the missing frequencies x and y if the sum of the frequencies is 100. Class interval 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 7 10 x 13 y 10 14 9

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Solution

## The given data is shown as follows: Class interval Frequency (fi) Class mark (xi) fixi 0−10 7 5 35 10−20 10 15 150 20−30 x 25 25x 30−40 13 35 455 40−50 y 45 45y 50−60 10 55 550 60−70 14 65 910 70−80 9 75 675 Total ∑ fi = 63 + x + y ∑ fixi = 2775 + 25x + 45y Sum of the frequencies = 100 $⇒\sum _{i}{f}_{i}=100\phantom{\rule{0ex}{0ex}}⇒63+x+y=100\phantom{\rule{0ex}{0ex}}⇒x+y=100-63\phantom{\rule{0ex}{0ex}}⇒x+y=37\phantom{\rule{0ex}{0ex}}⇒y=37-x....\left(1\right)$ Now, The mean of given data is given by $\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒42=\frac{2775+25x+45y}{100}\phantom{\rule{0ex}{0ex}}⇒4200=2775+25x+45y\phantom{\rule{0ex}{0ex}}⇒4200-2775=25x+45y\phantom{\rule{0ex}{0ex}}⇒1425=25x+45\left(37-x\right)\left[\mathrm{from}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}⇒1425=25x+1665-45x\phantom{\rule{0ex}{0ex}}⇒20x=1665-1425\phantom{\rule{0ex}{0ex}}⇒20x=240\phantom{\rule{0ex}{0ex}}⇒x=12$ If x = 12, then y = 37 − 12 = 25 ​Thus, the value of x is 12 and y is 25.

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