Statistics
Mean deviation about mean
nta_pyq_2023_jan
Grade 11
Question:
Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \ldots, a_{100}$ is 25. Then $S$ is
$\phi$
$\{99\}$
$\mathbb{N}$
$\{9\}$
Step-by-Step Solution
Key Concept: For 100 consecutive integers starting at $a_1$: mean $= a_1 + \frac{99}{2}$. Mean deviation $= \frac{\sum|x_i - \bar{x}|}{100}$. By symmetry this equals $\frac{2(\frac{1}{2}+\frac{3}{2}+\ldots+\frac{99}{2})}{100}$.
Mean deviation $= \frac{1+3+5+\ldots+99}{100} = \frac{50^2}{100} = 25$ for any $a_1 \in \mathbb{N}$. So $S = \mathbb{N}$.
Correct Answer: 3