Probability Distributions
DAILY_CHALLENGE
Grade None
Question:
Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x,P(X=x))$, $x=0,1,2,3,4$, lie on a fixed straight line in the $xy$-plane, and $P(X=x)=0$ for all $x\in\mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\dfrac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24\alpha$ is ___.
Step-by-Step Solution
Key Concept: Probabilities on a line give two linear equations in slope and intercept; then compute variance
Let $P(X=x)=mx+c$. Conditions:
$\sum_{x=0}^{4}(mx+c)=10m+5c=1$ and $\sum_{x=0}^{4}x(mx+c)=30m+10c=\dfrac{5}{2}$.
From first: $c=\dfrac{1-10m}{5}$. Substitute: $30m+2(1-10m)=\dfrac{5}{2}\Rightarrow10m=\dfrac{1}{2}\Rightarrow m=\dfrac{1}{20}$, $c=\dfrac{1}{10}$.
$P(X=x)=\dfrac{x+2}{20}$ for $x=0,1,2,3,4$.
$E[X^2]=\sum x^2\cdot\dfrac{x+2}{20}=\dfrac{0+3+16+45+96}{20}=\dfrac{160}{20}=8$.
$\alpha=E[X^2]-(E[X])^2=8-\dfrac{25}{4}=\dfrac{7}{4}$.
$24\alpha=24\cdot\dfrac{7}{4}=42$.
Correct Answer: 42