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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let the mean and variance of $8$ numbers
$-10,; -7,; -1,; x,; y,; 9,; 2,; 16$
be $\dfrac{7}{2}$ and $\dfrac{293}{4}$, respectively. Then the mean of $4$ numbers $x,; y,; x+y+1,; |x-y|$ is:
Solution
Mean $\dfrac{-18 + x + y + 2 + 9 + 16}{8} = \dfrac{7}{2}$ $\dfrac{x + y + 9}{8} = \dfrac{7}{2}$ $x + y + 9 = 28 \Rightarrow x + y = 19 \quad ...(1)$ Variance $\dfrac{\sum x_i^2}{8} - \mu^2 = \dfrac{293}{4}$ $\dfrac{10^2 + 7^2 + 1^2 + x^2 + y^2 + 9^2 + 2^2 + 16^2}{8} - \left(\dfrac{7}{2}\right)^2 = \dfrac{293}{4}$ Solving (1) & (2) ⇒ $x = 12,; y = 7$ Mean of required numbers $\dfrac{x + y + (x+y+1) + |x-y|}{4}$ $= \dfrac{12 + 7 + 20 + 5}{4} = \dfrac{44}{4} = 11$Online Test Series, Information About Examination,
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