What is the mean of $f (x) = 3 x + 2$ where x is a random variable with probability distribution.
$X = x$ $1$ $2$ $3$ $4$
$P (X = x)$ $1_{/ 6}$ $1_{/ 3}$ $1_{/ 3}$ $1_{/ 6}$

\frac{5}{2}

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\frac{15}{2}

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\frac{5}{3}

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\frac{19}{2}

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Solution

The correct option is D $\frac{19}{2}$
Given, $f (x) = 3 x + 2$

applying expectation on both sides, we get

$E (f (x)) = 3 E (x) + 2$ (by the properties of expectation)

mean $E (x) = \sum_{i = 1}^{n} x_{i} P (x_{i})$

Given, number of samples $n = 4$

Hence, mean = $\sum_{i = 1}^{4} x_{i} P (x_{i})$

substituting the values from the table we get,

$= 1 (\frac{1}{6}) + 2 (\frac{1}{3}) + 3 (\frac{1}{3}) + 4 (\frac{1}{6})$

$= \frac{1 + 4 + 6 + 4}{6} = \frac{15}{6}$

$E (x) = \frac{5}{2}$

Therefore, $E (f (x)) = 3 (\frac{5}{2}) + 2$

$= \frac{19}{2}$

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\begin{matrix} X = x & 1 & 2 & 3 & 4 P (X = x) & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6} \end{matrix}

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$X = x$	$1$	$2$	$3$	$4$
$P (X = x)$	$1_{/ 6}$	$1_{/ 3}$	$1_{/ 3}$	$1_{/ 6}$