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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are $ 2, 3, 5, 10, 11, 13, 15, 21 $, then the mean deviation about the median of all the 10 observations is
Solution
$ \frac{2+3+5+10+11+13+15+21+a+b}{10} = 9 $
$ \Rightarrow \frac{80 + a + b}{10} = 9 \Rightarrow a + b = 10 \quad ...(1)$
$ \frac{\sum x_i^2}{10} - \left( \frac{\sum x_i}{10} \right)^2 = 34.2 $
$ \Rightarrow \frac{2^2 + 3^2 + 5^2 + 10^2 + 11^2 + 13^2 + 15^2 + 21^2 + a^2 + b^2}{10} - 81 = 34.2 $
$ \Rightarrow 1094 + a^2 + b^2 - 810 = 342 $
$ \Rightarrow a^2 + b^2 = 58 \quad ...(2)$
From (1) and (2):
$ a = 7,; b = 3 $ or $ a = 3,; b = 7 $
Number $ \Rightarrow 2, 3, 5, 7, 10, 11, 13, 15, 21 $
Mean $ = \frac{7 + 10}{2} = 8.5 $
M.D. $ = \frac{6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5}{10} = \frac{50}{10} = 5 $
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