Discrete probability distributions

FS1-1 Discrete probability distributions — coverage pack

1 specification leaves · notes, questions, answers and worked methods

FS1-1.1 · Calculation of the mean and variance of discrete probability distributions. Extension of expected value function to include E(g(X)).

For a discrete random variable, $E(X)=\sum xP(X=x)$ and $\operatorname{Var}(X)=E(X^2)-[E(X)]^2$ .
For a function of the variable, calculate $E[g(X)]=\sum g(x)P(X=x)$ over every possible value of $X$ .
For example, once $E(X)$ and $E(X^2)$ are known, $E[(X-a)^2]=E(X^2)-2aE(X)+a^2$ can be evaluated without rebuilding a table.
A common error is to write $E[g(X)]=g(E(X))$ ; this is not true in general.

Tier 1 · Easy

1. The random variable $X$ takes values $0,1,2,4$ with probabilities $0.15,0.25,0.35,0.25$ respectively. Find $E(X)$ and $\operatorname{Var}(X)$ .[4 marks]
Answer
- $E(X)=1.95$
- $\operatorname{Var}(X)=1.8475$
Method: $E(X)=0(0.15)+1(0.25)+2(0.35)+4(0.25)=1.95$ . Also $E(X^2)=0+1(0.25)+4(0.35)+16(0.25)=5.65$ . Hence $\operatorname{Var}(X)=5.65-1.95^2=1.8475$ .

Tier 2 · Standard

1. The random variable $X$ takes values $-2,1,3$ with probabilities $0.2,0.5,0.3$ respectively. Calculate $E[(X+2)^2]$ .[3 marks]
Answer
- $E[(X+2)^2]=12$
Method: Evaluate the function at each support value: $(X+2)^2$ is $0,9,25$ . Therefore $E[(X+2)^2]=0(0.2)+9(0.5)+25(0.3)=12$ .

Tier 3 · Hard

1. The random variable $X$ takes values $0,1,3,5$ with probabilities $k,2k,3k,4k$ respectively. Find $k$ , $E(X)$ , $\operatorname{Var}(X)$ and $E[(X-2)^2]$ .[7 marks]
Answer
- $k=0.1$
- $E(X)=3.1$
- $\operatorname{Var}(X)=3.29$
- $E[(X-2)^2]=4.5$
Method: $10k=1$ , so $k=0.1$ . Then $E(X)=0+1(0.2)+3(0.3)+5(0.4)=3.1$ and $E(X^2)=1(0.2)+9(0.3)+25(0.4)=12.9$ . Thus $\operatorname{Var}(X)=12.9-3.1^2=3.29$ . Finally, $E[(X-2)^2]=E(X^2)-4E(X)+4=12.9-12.4+4=4.5$ .