Date | November 2016 | Marks available | 10 | Reference code | 16N.3sp.hl.TZ0.4 |
Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
Command term | Prove that | Question number | 4 | Adapted from | N/A |
Question
Two independent discrete random variables X and Y have probability generating functions G(t) and H(t) respectively. Let Z=X+Y have probability generating function J(t).
Write down an expression for J(t) in terms of G(t) and H(t).
By differentiating J(t), prove that
(i) E(Z)=E(X)+E(Y);
(ii) Var(Z)=Var(X)+Var(Y).
Markscheme
J(t)=G(t)H(t) A1
[1 mark]
(i) J′(t)=G′(t)H(t)+G(t)H′(t) M1A1
J′(1)=G′(1)H(1)+G(1)H′(1) M1
J′(1)=G′(1)+H′(1) A1
so E(Z)=E(X)+E(Y) AG
(ii) J″ M1A1
J''(1) = G''(1)H(1) + 2G'(1)H'(1) + G(1)H''(1)
= G''(1) + 2G'(1)H'(1) + H''(1) A1
{\text{Var}}(Z) = J''(1) + J'(1) - {\left( {J'(1)} \right)^2} M1
= G''(1) + 2G'(1)H'(1) + H''(1) + G'(1) + H'(1) - {\left( {G'(1) + H'(1)} \right)^2} A1
= G''(1) + G'(1) - {\left( {G'(1)} \right)^2} + H''(1) + H'(1) - {\left( {H'(1)} \right)^2} A1
so {\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y) AG
Note: If addition is wrongly used instead of multiplication in (a) it is inappropriate to give FT apart from the second M marks in each part, as the working is too simple.
[10 marks]