Question

The following table gives the probability distribution of a random variable \(X\): \[ \begin{array}{|c|c|c|c|c|} \hline x_i & 0 & 1 & 2 & 3 \\ \hline P(X=x_i) & 0.1 & 0.3 & 0.5 & 0.1 \\ \hline \end{array} \] The standard deviation of \(X\) is:

• A. \(0.4\)
• B. \(0.42\)
• C. \(0.8\)
• D. \(0.64\)

Hint:

For discrete probability distributions: \[ E(X)=\sum x_i p_i \] and \[ \sigma=\sqrt{E(X^2)-[E(X)]^2} \] Always calculate the mean first before computing variance and standard deviation.

Answer 1

The Correct Option is C

Concept: For a discrete random variable: Mean: \[ \mu=E(X)=\sum x_iP(X=x_i) \] Variance: \[ \sigma^2=E(X^2)-[E(X)]^2 \] Standard deviation: \[ \sigma=\sqrt{\sigma^2} \] We now compute each quantity carefully.

Step 1: Calculate the mean \(E(X)\).
\[ E(X) = 0(0.1)+1(0.3)+2(0.5)+3(0.1) \] \[ = 0+0.3+1+0.3 \] \[ = 1.6 \] Thus, \[ \mu=1.6 \]

Step 2: Calculate \(E(X^2)\).
\[ E(X^2) = 0^2(0.1)+1^2(0.3)+2^2(0.5)+3^2(0.1) \] \[ = 0+0.3+4(0.5)+9(0.1) \] \[ = 0.3+2+0.9 \] \[ = 3.2 \]

Step 3: Compute variance.
\[ \sigma^2 = E(X^2)-[E(X)]^2 \] \[ = 3.2-(1.6)^2 \] \[ = 3.2-2.56 \] \[ = 0.64 \]

Step 4: Find standard deviation.
\[ \sigma=\sqrt{0.64} \] \[ =0.8 \] Hence, \[ \boxed{0.8} \] Therefore the correct option is: \[ \boxed{(C)} \]