| Date | November 2016 | Marks available | 2 | Reference code | 16N.2.hl.TZ0.1 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Determine | 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 \({\text{E}}({X^2})\).
Find the value of \({\text{Var}}(X)\).
Markscheme
\({\text{E}}({X^2}) = \Sigma {x^2} \bullet {\text{P}}(X = x) = 10.37{\text{ }}( = 10.4{\text{ 3 sf)}}\) (M1)A1
[2 marks]
METHOD 1
\({\text{sd}}(X) = 1.44069 \ldots \) (M1)(A1)
\({\text{Var}}(X) = 2.08{\text{ }}( = 2.0756)\) A1
METHOD 2
\({\text{E}}(X) = 2.88{\text{ }}( = 0.06 + 0.27 + 0.5 + 0.98 + 0.63 + 0.44)\) (A1)
use of \({\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2}\) (M1)
Note: Award (M1) only if \({\left( {{\text{E}}(X)} \right)^2}\) is used correctly.
\(\left( {{\text{Var}}(X) = 10.37 - 8.29} \right)\)
\({\text{Var}}(X) = 2.08{\text{ }}( = 2.0756)\) A1
Note: Accept 2.11.
METHOD 3
\({\text{E}}(X) = 2.88{\text{ }}( = 0.06 + 0.27 + 0.5 + 0.98 + 0.63 + 0.44)\) (A1)
use of \({\text{Var}}(X) = {\text{E}}\left( {{{\left( {X - {\text{E}}(X)} \right)}^2}} \right)\) (M1)
\((0.679728 + \ldots + 0.549152)\)
\({\text{Var}}(E) = 2.08{\text{ }}( = 2.0756)\) A1
[3 marks]
