Question

The following table shows the probability distribution for the random variable B.

           

b

1

10

q

101

0.2

0.4

0.2

0.2

 

(i)        Given that , show that  and solve this equation.                                                                                         (4 marks)

(ii)       Calculate the value of .                                                              (2 marks)

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} y & 1 & 10 & q & 101 \\ \hline P(Y=y) & 0.2 & 0.4 & 0.2 & 0.2 \\ \end{array}$$

(i) Given $Var(Y)=1385.2$

$$E(Y)=0.2(1)+0.4(10)+0.2(q)+0.2(101)$$

$$=24.4+0.2q$$

$$E(Y^2)=0.2(1)^2+0.4(10)^2+0.2(q)^2+0.2(101)^2$$

$$=2080.4+0.2q^2$$

$$Var(Y)=E(Y^2)-(E(Y))^2$$

$$=2080.4+0.2q^2-(24.4+0.2q)^2$$

$$=2080.4+0.2q^2-595.36-9.76q+0.04q^2$$

$$=0.16q^2-9.76q+1485.04$$

Then

$$0.16q^2-9.76q+1485.04=1385.2$$

$$0.24q^2-9.76q+99.84=0$$

$$q^2-61q+624=0, True$$

(ii)

$$D=(-61)^2-4(1)(624)=35^2$$

$$q=\dfrac{61\pm35}{2(1)}$$

$$q_1=\dfrac{61-35}{2}=13$$

$$q_2=\dfrac{61+35}{2}=48$$

$$E(Y)=24.4+13=37.4$$

Or

$$E(Y)=24.4+48=72.4$$

The greatest value of $E(Y)$ is 72.4.