For the following probability distribution. X 1 2 3 4 P(X) 10 5 10 5 E ( X 2 ) is equal to

E ( X 2 ) = P(X) = 1 10 + 4 5 + 9 10 + 16 5 = 10 + 5 + 10 + 5 = 10 = 10

Probability — Class 12 Maths Solution

exemplar objective MCQ NCERT,Exemp,Q.89,Page.284

Question

For the following probability distribution.
X 1 2 3 4
P(X) $\frac{1}{{10}}$ $\frac{1}{5}$ $\frac{3}{{10}}$ $\frac{2}{5}$
$E\left( {{X^2}} \right)$ is equal to

(a) 3
(b) 5
(c) 7
(d) 10 ✓ Correct

Step-by-step Solution

$E\left( {{X^2}} \right) = \Sigma {X^2}P(X) = 1 \cdot \frac{1}{{10}} + 4 \cdot \frac{1}{5} + 9 \cdot \frac{3}{{10}} + 16 \cdot \frac{2}{5}$

$= \frac{1}{{10}} + \frac{4}{5} + \frac{{27}}{{10}} + \frac{{32}}{5}$

$= \frac{{1 + 8 + 27 + 64}}{{10}} = 10$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Probability. Curated by Sachin Sharma. Free for all students.