A random variable X has the following probability distribution: X = x i - 2 - 1 0 1 2 P X = x i 1 6 k 1 4 k 1 6 The variance of this random variable is

Question

A random variable X has the following probability distribution: X = x i - 2 - 1 0 1 2 P X = x i 1 6 k 1 4 k 1 6 The variance of this random variable is, 0, 5 24, 3 24, 7 4

MARKS App · Accepted Answer

A random variable X has the following probability distribution: X = x i - 2 - 1 0 1 2 P X = x i 1 6 k 1 4 k 1 6 The given distribution is the probability distribution. ∴ 1 6 + k + 1 4 + k + 1 6 = 1 ⇒ k = 5 24 Now, x i P x = x i x i P x i x i 2 P x i - 2 1 6 - 1 3 2 3 - 1 k - k k 0 1 4 0 0 1 k k k 2 1 6 1 3 2 3 ∑ x i P x i = 0 ∑ x i 2 P x i = 4 3 + 2 k Variance = ∑ x i 2 P x i - ∑ x i P x i 2 . = 4 3 + 2 k - 0 = 4 3 + 2 × 5 24 = 7 4

$X = x_{i}$	$- 2$	$- 1$	$0$	$1$	$2$
$P (X = x_{i})$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

$X = x_{i}$	$- 2$	$- 1$	$0$	$1$	$2$
$P (X = x_{i})$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

$x_{i}$	$P (x = x_{i})$	$x_{i} P (x_{i})$	$x_{i}^{2} P (x_{i})$
$- 2$	$\frac{1}{6}$	$- \frac{1}{3}$	$\frac{2}{3}$
$- 1$	$k$	$- k$	$k$
$0$	$\frac{1}{4}$	$0$	$0$
$1$	$k$	$k$	$k$
$2$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{2}{3}$
		$\sum_{} x_{i} P (x_{i}) = 0$	$\sum_{} \{x_{i}^{2} P (x_{i})\} = (\frac{4}{3} + 2 k)$

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