A small class is surveyed, and the random variable X represents the number of pets owned by a randomly chosen student from the class. The probability distribution of X is shown in the table below, where k is a constant. x 0 1 2 3 P(X=x) 0.2 0.35 0.3 k (a) Find the value of k. [2…

A Level Maths 9709: E(X) and Var(X) from a Probability Distribution Table

$x$	0	1	2	3
$P(X=x)$	$0.2$	$0.35$	$0.3$	$k$

x

P(X=x)

0.2

0.35

0.3

k

Part (a): Finding the unknown probability

Since $X$ is a discrete random variable, all of its probabilities must sum to $1$ : $P(X=0)+P(X=1)+P(X=2)+P(X=3)=1$ $0.2+0.35+0.3+k=1$

Adding the known probabilities: $0.2+0.35+0.3=0.85$

So $0.85+k=1 \implies k=1-0.85=0.15$

Check: $0.2+0.35+0.3+0.15=1.00$ ✓, and every value lies between $0$ and $1$ , so this is a valid probability distribution.

Part (b): Finding E(X)

$E(X)=\sum xP(X=x) = 0(0.2)+1(0.35)+2(0.3)+3(0.15)$ $=0+0.35+0.6+0.45=1.4$

Part (c): Finding Var(X)

First find $E(X^2)=\sum x^2P(X=x)$ : $E(X^2)=0^2(0.2)+1^2(0.35)+2^2(0.3)+3^2(0.15)$ $=0(0.2)+1(0.35)+4(0.3)+9(0.15)$ $=0+0.35+1.2+1.35=2.9$

Then apply $\mathrm{Var}(X)=E(X^2)-\bigl[E(X)\bigr]^2$ : $\mathrm{Var}(X)=2.9-(1.4)^2=2.9-1.96=0.94$

Check by the direct definition: $\mathrm{Var}(X)=\sum (x-E(X))^2P(X=x)$ $=(0-1.4)^2(0.2)+(1-1.4)^2(0.35)+(2-1.4)^2(0.3)+(3-1.4)^2(0.15)$ $=(1.96)(0.2)+(0.16)(0.35)+(0.36)(0.3)+(2.56)(0.15)$ $=0.392+0.056+0.108+0.384=0.94$ which agrees exactly with the value found above.

Final answers

(a) $k=\boxed{0.15}$
(b) $E(X)=\boxed{1.4}$
(c) $\mathrm{Var}(X)=\boxed{0.94}$