Date | November 2016 | Marks available | 3 | Reference code | 16N.2.hl.TZ0.1 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
A random variable X has a probability distribution given in the following table.
Determine the value of E(X2).
Find the value of Var(X).
Markscheme
E(X2)=Σx2∙P(X=x)=10.37 (=10.4 3 sf) (M1)A1
[2 marks]
METHOD 1
sd(X)=1.44069… (M1)(A1)
Var(X)=2.08 (=2.0756) A1
METHOD 2
E(X)=2.88 (=0.06+0.27+0.5+0.98+0.63+0.44) (A1)
use of Var(X)=E(X2)−(E(X))2 (M1)
Note: Award (M1) only if (E(X))2 is used correctly.
(Var(X)=10.37−8.29)
Var(X)=2.08 (=2.0756) A1
Note: Accept 2.11.
METHOD 3
E(X)=2.88 (=0.06+0.27+0.5+0.98+0.63+0.44) (A1)
use of Var(X)=E((X−E(X))2) (M1)
(0.679728+…+0.549152)
Var(E)=2.08 (=2.0756) A1
[3 marks]