A random variable X has the probability distribution given below. X 1 2 3 4 5 P(X = x) K 2K 3K 2K K Its variance is

Question

A random variable X has the probability distribution given below. X 1 2 3 4 5 P(X = x) K 2K 3K 2K K Its variance is (A) (16/3) (B) (4/3) (C) (5/3) (D) (10/3)

Tardigrade Admin · Accepted Answer

Given distribution is

X	1	2	3	4	5
P(X = x)	K	2K	3K	2K	K

∴

Variance

= Σ x_{i}^{2} p - (\sum x_{i} p)^{2}

= (1 k + 8 k + 27 k + 32 k + 25 k) - (k + 4 k + 9 k + 8 k + 5 k)^{2}

= (93 k) - (27 k)^{2} = (93 \times \frac{1}{9}) - (27 \times \frac{1}{<} b r / >< b r / > 9)^{2} (∵ Σ p = 1, ∴ k = \frac{1}{9})

= \frac{93}{9} - 9 = \frac{93 - 81}{9} = \frac{12}{9} = \frac{4}{3}

Q. A random variable $X$ has the probability distribution given below.

X 1 2 3 4 5

P(X = x) K 2K 3K 2K K

Its variance is

$\frac{16}{3}$

$\frac{4}{3}$

$\frac{5}{3}$

$\frac{10}{3}$

Q. A random variable X has the probability distribution given below. X 1 2 3 4 5 P(X = x) K 2K 3K 2K K Its variance is

316​

34​

35​

310​

Given distribution is X 1 2 3 4 5 P(X = x) K 2K 3K 2K K ∴ Variance =Σxi2​p−(∑xi​p)2 =(1k+8k+27k+32k+25k)−(k+4k+9k+8k+5k)2 =(93k)−(27k)2=(93×91​)−(27×<1​br/><br/>9)2(∵Σp=1,∴k=91​) =993​−9=993−81​=912​=34​

Q. A random variable $X$ has the probability distribution given below.

X 1 2 3 4 5

P(X = x) K 2K 3K 2K K

Its variance is

$\frac{16}{3}$

$\frac{4}{3}$

$\frac{5}{3}$

$\frac{10}{3}$