Porisma

A hint to get started

Normalise the parameterised distribution before transforming variance

The probability function of X is

P(X=x)={(2x−1)θ,x=1,2,3,k,x=4,0,otherwise,P(X=x)=\begin{cases}(2x-1)\theta,&x=1,2,3,\\k,&x=4,\\0,&\text{otherwise},\end{cases}

where 0<θ<1/90<\theta<1/9.

(i) Show that k=1−9θk=1-9\theta. Find the distribution of X in terms of θ.

(ii) Find E(X)E(X) in terms of θ and hence show that Var⁡(X)=26θ−196θ2\operatorname{Var}(X)=26\theta-196\theta^2.

(iii) Y=a+bXY=a+bX, with a,b non-zero constants. Given Var⁡(Y)=b2/3\operatorname{Var}(Y)=b^2/3, find θ.

[2017/MJC/II/6]

CJC · Discussion · Question 5

A constant shift changes the mean but not the variance; a scale factor contributes its square.

Read the worked solution