Let the mean and the standard deviation of the probability distribution \(x:\alpha,1,0,-3\); \(P(x):\frac{1}{3},K,\frac{1}{6},\frac{1}{4}\) be \(\mu\) and \(\sigma\), respectively. If \(\sigma-\mu=2\), then \(\sigma+\mu\) is equal to ____ .

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Statistics

Let the mean and the standard deviation of the probability distribution

x : α, 1, 0, - 3

;

P (x) : \frac{1}{3}, K, \frac{1}{6}, \frac{1}{4}

μ

and

σ

, respectively. If

σ - μ = 2

, then

σ + μ

is equal to ____ .

Answer: 5

Probabilities sum to 1:

\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1 ⟹ K = \frac{1}{4}

μ = α (\frac{1}{3}) + 1 (\frac{1}{4}) + 0 + (- 3) (\frac{1}{4}) = \frac{α}{3} - \frac{1}{2}

E [X^{2}] = \frac{α ^{2}}{3} + \frac{1}{4} + 9 (\frac{1}{4}) = \frac{α ^{2}}{3} + \frac{5}{2}, σ^{2} = E [X^{2}] - μ^{2}

Imposing

σ - μ = 2

(i.e.

σ = μ + 2

) and squaring,

σ^{2} = μ^{2} + 4 μ + 4 = E [X^{2}] - μ^{2} \Rightarrow 2 μ^{2} + 4 μ + 4 = E [X^{2}]

, which reduces to a quadratic in

α

with roots

α = 0

(rejected, since

X = 0

is already a listed outcome) and

α = 6

.
With

α = 6

μ = \frac{6}{3} - \frac{1}{2} = \frac{3}{2}

;

σ^{2} = \frac{36}{3} + \frac{5}{2} - \frac{9}{4} = 12 + 2.5 - 2.25 = 12.25 ⟹ σ = 3.5

(and indeed

σ - μ = 3.5 - 1.5 = 2

, confirming consistency).

σ + μ = 3.5 + 1.5 = 5

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Integer answer