We are given the mean and variance of five observations $\{1, 3, a, 7, b\}$ and asked to find the variance of a new set of observations where each term is modified by its index $n$.
Step 1: Find the values of $a$ and $b$
Using the definition of the Mean ($\bar{x}$):
$\bar{x} = \frac{1 + 3 + a + 7 + b}{5} = 5$
$11 + a + b = 25 \implies a + b = 14$ ... (Equation 1)
Using the definition of Variance ($\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$):
$10 = \frac{1^2 + 3^2 + a^2 + 7^2 + b^2}{5} - 5^2$
$10 = \frac{1 + 9 + a^2 + 49 + b^2}{5} - 25$
$35 = \frac{59 + a^2 + b^2}{5} \implies a^2 + b^2 + 59 = 175 \implies a^2 + b^2 = 116$ ... (Equation 2)
From Eq 1, $b = 14 - a$. Substitute into Eq 2:
$a^2 + (14 - a)^2 = 116 \implies a^2 + 196 + a^2 - 28a = 116$
$2a^2 - 28a + 80 = 0 \implies a^2 - 14a + 40 = 0$
$(a - 10)(a - 4) = 0$. Given $a > b$, we have **$a = 10$** and **$b = 4$**.
Step 2: Determine the new observations
The new observations are $y_n = n + x_n$ for $n = 1, 2, 3, 4, 5$:
- $y_1 = 1 + x_1 = 1 + 1 = 2$
- $y_2 = 2 + x_2 = 2 + 3 = 5$
- $y_3 = 3 + x_3 = 3 + 10 = 13$
- $y_4 = 4 + x_4 = 4 + 7 = 11$
- $y_5 = 5 + x_5 = 5 + 4 = 9$
The new data set is $\{2, 5, 13, 11, 9\}$.
Step 3: Calculate the Variance of the new set
Find the new Mean ($\bar{y}$):
$\bar{y} = \frac{2 + 5 + 13 + 11 + 9}{5} = \frac{40}{5} = 8$
Find the new Variance ($\sigma_y^2$):
$\sigma_y^2 = \frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5} - 8^2$
$\sigma_y^2 = \frac{4 + 25 + 169 + 121 + 81}{5} - 64$
$\sigma_y^2 = \frac{400}{5} - 64 = 80 - 64 = 16$
Similar Questions
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Important Questions from Measures of Dispersion and Probability
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x 0-10 10-20 20-30 30-40 40-50 f 3 6 2 x y $\Sigma f = 20$ are equal, then $xy^2$ is equal to
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