Question ID: #1159
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 $n+x_{n}$, $n=1,2,3,4,5$ is
- (1) 17
- (2) 16.4
- (3) 17.4
- (4) 16
Solution:
$$\overline{x} = \frac{\sum x_{i}}{n}$$
$$5 = \frac{1 + 3 + a + 7 + b}{5}$$
$$25 = 11 + a + b$$
$$a + b = 14$$
$$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} – (\overline{x})^{2}$$
$$10 = \frac{1^2 + 3^2 + a^2 + 7^2 + b^2}{5} – 5^2$$
$$10 = \frac{1 + 9 + a^2 + 49 + b^2}{5} – 25$$
$$35 = \frac{59 + a^2 + b^2}{5}$$
$$175 = 59 + a^2 + b^2$$
$$a^2 + b^2 = 116$$
$$(a+b)^2 – 2ab = 116$$
$$(14)^2 – 2ab = 116$$
$$196 – 2ab = 116$$
$$2ab = 80 \Rightarrow ab = 40$$
$$a = 10, b = 4 \quad (\text{since } a > b)$$
$$x_n = \{1, 3, 10, 7, 4\}$$
$$y_n = n + x_n = \{1+1, 2+3, 3+10, 4+7, 5+4\}$$
$$y_n = \{2, 5, 13, 11, 9\}$$
$$\overline{y} = \frac{2 + 5 + 13 + 11 + 9}{5}$$
$$\overline{y} = \frac{40}{5} = 8$$
$$\sigma_y^2 = \frac{\sum y_i^2}{n} – (\overline{y})^2$$
$$\sigma_y^2 = \frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5} – (8)^2$$
$$\sigma_y^2 = \frac{4 + 25 + 169 + 121 + 81}{5} – 64$$
$$\sigma_y^2 = \frac{400}{5} – 64$$
$$\sigma_y^2 = 80 – 64$$
$$\sigma_y^2 = 16$$
Ans. (4)
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