Let X be a discrete random variable. The probability distribution of X is given below:

X:
30 10 –10

P(X):
$\frac{1}{5}$ $\frac{3}{10}$ $\frac{1}{2}$

Then E(X) is equal to
(a) 6
(b) 4
(c) 3
(d) –5

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Solution

For a discrete random variable X,

X
P(X)

30
$\frac{1}{5}$

10
$\frac{3}{10}$

–10
$\frac{1}{2}$

$Since E (X) = \sum_{i = 1}^{n} x_{i} P (x_{i}) = 30 \times \frac{1}{5} + 10 \times \frac{3}{10} + (- 10) \times \frac{1}{2} = 6 + 3 - 5 = 9 - 5 i . e . E (X) = 4$

Hence, the correct answer is option B.

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