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$- a$ Let the mean and variance of four numbers 3, 7, x and y $(x > y)$ be 5 and 10 respectively. Then the mean of four numbers $3 + 2 x, 7 + 2 y, x + y$ and $x - y$ is 10

Let in a series of 2n observations, half of them are equal to a and remaining half are equal to -a. Also by adding a constant b to each of these observations, the mean and standard deviation of new set becomes 5 and 20, respectively. Then the value of $a^{2} + b^{2}$ is equal to 425

Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to 68

Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then $b^{2} = 3 (a^{2} + c^{2}) - 9 d^{2}$

answer is B, C, D.

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Detailed Solution

$5 = \frac{3 + 7 + x + y}{4} \Rightarrow x + y = 10$ $Var (x) = 10 = \frac{3^{2} + 7^{2} + x^{2} + y^{2}}{4} - 25$

$\begin{aligned} 140 = 49 + 9 + x^{2} + y^{2} \\ x^{2} + y^{2} = 82 \\ x + y = 10 \\ \Rightarrow (x, y) = (9, 1) \end{aligned} Mean = \frac{48}{4} = 12$

$\begin{aligned} \bar{x} = \frac{\sum x_{i}}{2 n} = \frac{(a + a + \dots + a) - (a + a + \dots + a)}{2 n} \\ \Rightarrow \bar{x} = 0 \end{aligned} and σ_{x}^{2} = \frac{\sum x_{i}^{2}}{2 n} - (\bar{x})^{2} = \frac{a^{2} + a^{2} + \dots + a^{2}}{2 n} - 0 = a^{2}$

$\Rightarrow σ_{x} = a$

$Now, adding a constant b then \bar{y} = \bar{x} + b = 5 and σ_{y} = σ_{x} (No change in S.D.) \Rightarrow a = 20 \Rightarrow a^{2} + b^{2} = 425$

$Let number be a_{1}, a_{2}, a_{3}, \dots \dots . a_{2 n}, b_{1}, b_{2}, b_{3} \dots b_{n} \begin{aligned} σ^{2} = \frac{\sum a^{2} + \sum b^{2}}{3 n} - (5)^{2} \\ \Rightarrow \sum a^{2} + \sum b^{2} = 87 n \end{aligned}$

The distribution is

$a_{1} + 1, a_{2} + 1, a_{3} + 1, \dots \dots . a_{2 n} + 1, b_{1} - 1, v a r i a n c e = \begin{aligned} = \frac{\sum (a + 1)^{2} + \sum (b - 1)^{2}}{3 n} - {(\frac{12 n + 2 n + 3 n - n}{3 n})}^{2} \\ = \frac{(\sum a^{2} + 2 n + 2 \sum a) + (\sum b^{2} + n - 2 \sum b)}{3 n} \\ = \frac{(\sum a^{2} + 2 n + 2 \sum a) + (\sum b^{2} + n - 2 \sum b)}{3 n} - {(\frac{16}{3})}^{2} \\ = \frac{87 n + 3 n + 2 (12 n) - 2 (3 n)}{3 n} - {(\frac{16}{3})}^{2} \\ \Rightarrow k = \frac{108}{3} - {(\frac{16}{5})}^{2} \\ \Rightarrow 9 k = 3 (108) - (16)^{2} = 324 - 256 = 68 \end{aligned}$

$F o r t h r e e n u m b e r s a, b, c mean = \frac{a + b + c}{3} (= \bar{x}) \begin{aligned} b = a + c \\ \Rightarrow \bar{x} = \frac{2 b}{3} \\ S.D. (a + 2, b + 2, c + 2) \\ = S.D. (a, b, c) = d \\ \Rightarrow d^{2} = \frac{a^{2} + b^{2} + c^{2}}{3} - (\bar{x})^{2} \\ \Rightarrow d^{2} = \frac{a^{2} + b^{2} + c^{2}}{3} - \frac{4 b^{2}}{9} \\ \Rightarrow 9 d^{2} = 3 (a^{2} + b^{2} + c^{2}) - 4 b^{2} \\ \Rightarrow b^{2} = 3 (a^{2} + c^{2}) - 9 d^{2} \end{aligned}$