Let and Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, a,b,c be 18, 4, and , respectively. Then is equal to

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Solution

Q.
Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

Ans.
(33)

$a, b, c \in N$ and $a < b < c$

Since, mean = 18 $\Rightarrow \frac{9 + 25 + a + b + c}{5} = 18$

$\Rightarrow 34 + a + b + c = 90 \Rightarrow a + b + c = 56$

Mean deviation about the mean is 4

$\Rightarrow \frac{| 9 - 18 | + | 25 - 18 | + | a - 18 | + | b - 18 | + | c - 18 |}{5} = 4$

$\Rightarrow 9 + 7 + | a - 18 | + | b - 18 | + | c - 18 | = 20$

$\Rightarrow | a - 18 | + | b - 18 | + | c - 18 | = 4$

Variance $= \frac{136}{5}$

$\Rightarrow \frac{\sum | x_{i} - \bar{x} |^{2}}{n} = \frac{136}{5}$

$\Rightarrow \frac{81 + 49 + | a - 18 |^{2} + | b - 18 |^{2} + | c - 18 |^{2}}{5} = \frac{136}{5}$

$\Rightarrow | a - 18 |^{2} + | b - 18 |^{2} + | c - 18 |^{2} = 136 - 130 = 6$

Possible values are $| a - 18 |^{2} = 1, | b - 18 |^{2} = 1$ and $| c - 18 |^{2} = 4$

$∵$ $a < b < c$

$∴$ $18 - a = 1, b - 18 = 1$ and $c - 18 = 2$

$\Rightarrow a = 17, b = 19$ and $c = 20$

$Hence, 2 a + b - c = 34 + 19 - 20 = 33$

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