Probability
Probability Distribution — Mean and Standard Deviation
nta_pyq_2024_apr
Grade 12
Question:
Let the mean and the standard deviation of the probability distribution $\begin{array}{|c|c|c|c|c|}\hline X & \alpha & 1 & 0 & -3\\\hline P(X) & \frac{1}{3} & K & \frac{1}{6} & \frac{1}{4}\\\hline\end{array}$ be $\mu$ and $\sigma$, respectively. If $\sigma-\mu=2$, then $\sigma+\mu$ is equal to
Step-by-Step Solution
Key Concept: Sum of probabilities $=1$: $\frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1\Rightarrow K=\frac{1}{4}$. $\mu=\frac{\alpha}{3}+1\cdot K+0-\frac{3}{4}=\frac{\alpha}{3}-\frac{1}{2}$. Given $\sigma=\mu+2$, solve for $\alpha$.
$\alpha=6$, $\mu=3/2$, $\sigma=7/2$. $\sigma+\mu=5$.
Correct Answer: 5