Statistics
Variance of an AP
nta_pyq_2023_jan
Grade None
Question:
Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in A.P. and $a_1 + a_3 = 10$. If the mean of these six numbers is $\dfrac{19}{2}$ and their variance is $\sigma^2$, then $8\sigma^2$ is equal to
Step-by-Step Solution
Key Concept: From $a_1 + a_3 = 10$: $2a_1 + 2d = 10 \Rightarrow a_1 + d = 5$. From mean $= \frac{19}{2}$: $a_1 + a_6 = 19$. Solve for $a_1$ and $d$, then compute variance using $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$.
$a_1 = 2$, $d = 3$. Numbers: 2, 5, 8, 11, 14, 17. $\sigma^2 = \frac{2^2+5^2+8^2+11^2+14^2+17^2}{6} - \left(\frac{19}{2}\right)^2 = \frac{699}{6} - \frac{361}{4} = \frac{105}{4}$. $8\sigma^2 = 210$.
Correct Answer: 2