CAIE S2 2015 June — Question 7 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypePure expectation and variance calculation
DifficultyModerate -0.8 This question tests standard application of expectation and variance rules for linear combinations of independent random variables, requiring only direct formula substitution with known distributions B(10,0.8) and Po(3). Part (iii) adds mild complexity by requiring enumeration of cases where 2X - Y = 18, but all components are routine A-level techniques with no problem-solving insight needed.
Spec5.02d Binomial: mean np and variance np(1-p)5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY)

7 The independent variables \(X\) and \(Y\) are such that \(X \sim \mathrm {~B} ( 10,0.8 )\) and \(Y \sim \mathrm { Po } ( 3 )\). Find
  1. \(\mathrm { E } ( 7 X + 5 Y - 2 )\),
  2. \(\operatorname { Var } ( 4 X - 3 Y + 3 )\),
  3. \(\mathrm { P } ( 2 X - Y = 18 )\).

Question 7:
Part (i):
AnswerMarks Guidance
\(7E(X) + 5E(Y) - 2\)M1 allow incorrect means
\((= 7 \times 8 + 5 \times 3) - 2\)
AnswerMarks
\(= 69\)A1 [2]
Part (ii):
AnswerMarks Guidance
\(\text{Var}(X) = 1.6, \text{Var}(Y) = 3\)B1 both
\(16\text{Var}(X) + 9\text{Var}(Y)\)M1 M1 for mult by 16 and 9; allow with \(`+ 3'\)
M1M1 for add without \(`+ 3'\); allow incorrect multipliers
\((= 16 \times 1.6 + 9 \times 3)\)
AnswerMarks
\(= 52.6\)A1 [4]
Part (iii):
AnswerMarks Guidance
\(X = 10, Y = 2\) and \(X = 9, Y = 0\)B1 both pairs seen or implied
\(0.8^{10} \times e^{-3} \times \dfrac{3^2}{2}\) or \(10 \times 0.8^9 \times 0.2 \times e^{-3}\)M1 or 0.0241 or 0.0134 (3sf) one correct product
\(0.8^{10} \times e^{-3} \times \dfrac{3^2}{2} + 10 \times 0.8^9 \times 0.2 \times e^{-3}\)M1
\(= 0.0374/5\)A1 [4] all correct
## Question 7:

### Part (i):
$7E(X) + 5E(Y) - 2$ | M1 | allow incorrect means
$(= 7 \times 8 + 5 \times 3) - 2$
$= 69$ | A1 [2] |

### Part (ii):
$\text{Var}(X) = 1.6, \text{Var}(Y) = 3$ | B1 | both
$16\text{Var}(X) + 9\text{Var}(Y)$ | M1 | M1 for mult by 16 and 9; allow with $`+ 3'$
| M1 | M1 for add without $`+ 3'$; allow incorrect multipliers
$(= 16 \times 1.6 + 9 \times 3)$
$= 52.6$ | A1 [4] |

### Part (iii):
$X = 10, Y = 2$ and $X = 9, Y = 0$ | B1 | both pairs seen or implied

$0.8^{10} \times e^{-3} \times \dfrac{3^2}{2}$ or $10 \times 0.8^9 \times 0.2 \times e^{-3}$ | M1 | or 0.0241 or 0.0134 (3sf) one correct product

$0.8^{10} \times e^{-3} \times \dfrac{3^2}{2} + 10 \times 0.8^9 \times 0.2 \times e^{-3}$ | M1 |
$= 0.0374/5$ | A1 [4] | all correct
7 The independent variables $X$ and $Y$ are such that $X \sim \mathrm {~B} ( 10,0.8 )$ and $Y \sim \mathrm { Po } ( 3 )$. Find\\
(i) $\mathrm { E } ( 7 X + 5 Y - 2 )$,\\
(ii) $\operatorname { Var } ( 4 X - 3 Y + 3 )$,\\
(iii) $\mathrm { P } ( 2 X - Y = 18 )$.

\hfill \mbox{\textit{CAIE S2 2015 Q7 [10]}}