Step 1: Formula for expected value.
The expected value \( E(X) \) of a discrete random variable is given by:
\[
E(X) = \sum x_i \cdot P(x_i)
\]
where \( x_i \) are the values of the random variable, and \( P(x_i) \) are the corresponding probabilities.
Step 2: Substitute the given values.
We have the following values for \( X \):
- \( x_1 = -1 \), \( P(x_1) = \frac{1}{8} \)
- \( x_2 = 0 \), \( P(x_2) = \frac{1}{2} \)
- \( x_3 = 2 \), \( P(x_3) = \frac{3}{8} \)
Substituting these into the formula:
\[
E(X) = (-1) \cdot \frac{1}{8} + 0 \cdot \frac{1}{2} + 2 \cdot \frac{3}{8}
\]
\[
E(X) = \frac{-1}{8} + 0 + \frac{6}{8}
\]
\[
E(X) = \frac{5}{8} = 0.625
\]
Step 3: Conclusion.
The expected value of \( X \) is 0.625, which is closest to option (C).
If the probability function for a random variable \( x \) is given as \( f(x) = \frac{x+3}{15} \) when \( x = 1, 2, 3 \), find the sum of the values of the probability distribution for \( x \).
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: