AHL 4.14—Properties of discrete and continuous random variables

For any valid probability distribution, all probabilities must sum to 1: ${\sum }_{n}P(N=n)=1$

Substituting the known probabilities: $0.1+0.2+p+0.25+0.15=1$

$0.7+p=1⟹p=0.30$

The value of $p$ is 0.30.

We need all probabilities to sum to exactly 1. The known values sum to $0.1+0.2+0.25+0.15=0.70$, so $p$ must make up the remaining $1-0.70=0.30$.

If $p=0.25$, the total would be $0.70+0.25=0.95\ne 1$. Eliminated.

If $p=0.20$, the total would be $0.70+0.20=0.90\ne 1$. Eliminated.

If $p=0.40$, the total would be $0.70+0.40=1.10>1$, which is impossible. Eliminated.

$p=0.30$ gives a total of $0.70+0.30=1.00$ ✓. The answer is 0.30.

Use the shortcut: $\text{Var}(S)=E(S^2)-[E(S){]}^2$, where $[E(S){]}^2=(2.6{)}^2=6.76$.

$E(S^2)=1^2(0.1)+2^2(0.4)+3^2(0.3)+4^2(0.2)$ $=0.1+1.6+2.7+3.2=7.6$

$\text{Var}(S)=7.6-(2.6{)}^2=7.6-6.76=0.84$

$\text{Var}(S)=0.84$.

We need $E(S^2)-[E(S){]}^2$. Compute $E(S^2)=0.1+1.6+2.7+3.2=7.6$.

$6.76=(2.6{)}^2=[E(S){]}^2$. This is the square of the mean, not the variance — a classic error. Eliminated.

There is no calculation path from this distribution that yields $\text{Var}(S)=1.00$; it does not equal $E(S^2)-[E(S){]}^2=7.6-6.76$. Eliminated.

$0.70$ would result from an arithmetic error in computing $E(S^2)$. Careful calculation gives $7.6-6.76=0.84\ne 0.70$. Eliminated.

$\text{Var}(S)=7.6-6.76=0.84$.