JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Statistics

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}, \dots\dots, a_{100}

is 25 . Then S is

Answer: (C)

Solution

For the

100

consecutive integers

a_{1}, a_{1} + 1, \dots, a_{1} + 99

, the mean is

a_{1} + \frac{99}{2}

, and the deviations from it are

\pm \frac{1}{2}, \pm \frac{3}{2}, \dots, \pm \frac{99}{2}

. The mean deviation is

\frac{2}{100} (\frac{1}{2} + \frac{3}{2} + \dots + \frac{99}{2}) = \frac{1 + 3 + \dots + 99}{100} = \frac{2500}{100} = 25

, a value free of

a_{1}

. Since the mean deviation is

25

for every positive integer

a_{1}

, the set

S

N

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option