Let the Mean and Variance of five observations and be 5 and 10 respectively. Then the Variance of the observations is [2025]

Question

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

Smart Achievers · Accepted Answer

(3)

We have, $\bar{x} = 5 and σ^{2} = 10$

Now, $\bar{x} = \frac{\sum_{}^{} x_{i}^{}}{n} = \frac{1 + 3 + a + 7 + b}{5} \Rightarrow \frac{11 + a + b}{5} = 5$

$\Rightarrow a + b = 14$

Also, $σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\bar{x})}^{2}$

$\Rightarrow 10 = \frac{1^{2} + 3^{2} + a^{2} + 7^{2} + b^{2}}{5} - 25$

$\Rightarrow a^{2} + b^{2} = 116$

Since, a > b $\Rightarrow$ a = 10 and b = 4

Now, $n + x_{n}$ : 2, 5, 13, 11, 9

$∴ σ^{2} = \frac{2^{2} + 5^{2} + 13^{2} + 11^{2} + 9^{2}}{5} - {(\frac{2 + 5 + 13 + 11 + 9}{5})}^{2} = 80 - 64 = 16$ .

Solution

Q.
Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

1 17

2 16.4

3 16

4 17.4

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