Let $9=\mathrm{x}_{1}<\mathrm{x}_{2}<\ldots<\mathrm{x}_{7}$ be in an A.P. with common difference d. If the standard deviation of $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots, \mathrm{x}_{7}$ is 4 and the mean is $\overline{\mathrm{x}}$, then $\overline{\mathrm{x}}+\mathrm{x}_{6}$ is equal to :

(A) $18\left(1+\frac{1}{\sqrt{3}}\right)$

(C) $2\left(9+\frac{8}{\sqrt{7}}\right)$

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Statistics

Let

9 = x_{1} < x_{2} < \dots < x_{7}

be in an A.P. with common difference d. If the standard deviation of

x_{1}, x_{2} \dots, x_{7}

is 4 and the mean is

\overline{x}

, then

\overline{x} + x_{6}

is equal to :

Answer: (B)

The seven terms

9, 9 + d, \dots, 9 + 6 d

have mean

\overset{x}{ˉ} = 9 + 3 d

. For an arithmetic progression of

n

terms the variance is

\frac{d ^{2} ( n ^{2} - 1 )}{12}

, which for

n = 7

equals

4 d^{2}

, so the standard deviation is

2∣ d ∣

. Setting

2∣ d ∣ = 4

gives

∣ d ∣ = 2

, and since the sequence increases,

d = 2

. Then

\overset{x}{ˉ} = 9 + 6 = 15

and

x_{6} = 9 + 5 (2) = 19

, so

\overset{x}{ˉ} + x_{6} = 34

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option