JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Statistics

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

\frac{19}{2}

and their variance is

σ^{2}

, then

8 σ^{2}

is equal to

Answer: (B)

Solution

In the A.P. the condition

a_{1} + a_{3} = 10

gives

2 (a_{1} + d) = 10

, so

a_{2} = a_{1} + d = 5

. A mean of

\frac{19}{2}

makes the total

57

, and

\frac{6}{2} (a_{1} + a_{6}) = 57

gives

a_{1} + a_{6} = 19

, i.e.

2 a_{1} + 5 d = 19

. With

a_{1} + d = 5

this yields

d = 3

and

a_{1} = 2

, so the numbers are

2, 5, 8, 11, 14, 17

. Their deviations from the mean

\frac{19}{2}

are

\pm \frac{15}{2}, \pm \frac{9}{2}, \pm \frac{3}{2}

, so

σ^{2} = \frac{2 ( \frac{225}{4} + \frac{81}{4} + \frac{9}{4} )}{6} = \frac{105}{4}

. Hence

8 σ^{2} = 210

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Single correct option