Statistics
Standard deviation of AP — mean and specific term
nta_pyq_2023_jan
Grade None
Question:
Let $9 = x_1 < x_2 < \ldots < x_7$ be in an A.P. with common difference $d$. If the standard deviation of $x_1, x_2, \ldots, x_7$ is 4 and the mean is $\bar{x}$, then $\bar{x} + x_6$ is equal to:
$18\left(1 + \dfrac{1}{\sqrt{3}}\right)$
34
$2\left(9 + \dfrac{8}{\sqrt{7}}\right)$
25
Step-by-Step Solution
Key Concept: The 7 terms are $9, 9+d, \ldots, 9+6d$. Mean $\bar{x} = 9 + 3d$. Variance $= \frac{1}{7}(0+1+4+9+16+25+36)d^2 - 9d^2 = \frac{91d^2}{7} - 9d^2 = 4d^2$. Set $\sigma = 4$.
$4d^2 = 16 \Rightarrow d = 2$. $\bar{x} = 15$, $x_6 = 19$. $\bar{x} + x_6 = 34$.
Correct Answer: 2