Date | May 2014 | Marks available | 16 | Reference code | 14M.3sp.hl.TZ0.4 |
Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
Command term | Find, Hence, Show that, and State | Question number | 4 | Adapted from | N/A |
Question
Consider the random variable X∼Geo(p)X∼Geo(p).
(a) State P(X<4)P(X<4).
(b) Show that the probability generating function for X is given by GX(t)=pt1−qtGX(t)=pt1−qt, where q=1−pq=1−p.
Let the random variable Y=2XY=2X.
(c) (i) Show that the probability generating function for Y is given by GY(t)=GX(t2)GY(t)=GX(t2).
(ii) By considering G′Y(1), show that E(Y)=2E(X).
Let the random variable W=2X+1.
(d) (i) Find the probability generating function for W in terms of the probability generating function of Y.
(ii) Hence, show that E(W)=2E(X)+1.
Markscheme
(a) use of P(X=n)=pqn−1 (q=1−p) (M1)
P(X<4)=p+pq+pq2 (=1−q3) (=1−(1−p)3) (=3p−3p2+p3) A1
[2 marks]
(b) GX(t)=P(X=1)t+P(X=2)t2+… (M1)
=pt+pqt2+pq2t3+… A1
summing an infinite geometric series M1
=pt1−qt AG
[3 marks]
(c) (i) EITHER
GY(t)=P(Y=1)t+P(Y=2)t2+… A1
=0×t+P(X=1)t2+0×t3+P(X=2)t4+… M1A1
=GX(t2) AG
OR
GY(t)=E(tY)=E(t2X) M1A1
=E((t2)X) A1
=GX(t2) AG
(ii) E(Y)=G′Y(1) A1
EITHER
=2tG′X(t2) evaluated at t=1 M1A1
=2E(X) AG
OR
=ddx(pt2(1−qt2))=2pt(1−qt2)+2pqt3(1−qt2)2 evaluated at t=1 A1
=2×p(1−qt)+pqt(1−qt)2 evaluated at t=1 (or 2p) A1
=2E(X) AG
[6 marks]
(d) (i) GW(t)=tGY(t) (or equivalent) A2
(ii) attempt to evaluate G′W(t) M1
EITHER
obtain 1×GY(t)+t×G′Y(t) A1
substitute t=1 to obtain 1×1+1×G′Y(1) A1
OR
=ddx(pt3(1−qt2))=3pt2(1−qt2)+2pqt4(1−qt2)2 A1
substitute t=1 to obtain 1+2p A1
=1+2E(X) AG
[5 marks]
Total [16 marks]