WJEC Further Unit 2 2023 June — Question 1 7 marks

Exam BoardWJEC
ModuleFurther Unit 2 (Further Unit 2)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Random Variables
TypeLinear combinations of independent variables
DifficultyModerate -0.8 This is a straightforward application of standard expectation and variance rules for linear combinations of independent random variables. Part (a) and (b) require only direct formula application (E[aX+bY+c] and Var[aX+bY+c]), while part (c) uses independence to write E[X²Y]=E[X²]E[Y], then applies Var(X)=E[X²]-E[X]². All three parts are routine calculations with no problem-solving or conceptual challenge, making this easier than average even for Further Maths.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)

The random variable \(X\) has mean 17 and variance 64. The independent random variable \(Y\) has mean 10 and variance 16. Find the value of
  1. E\((4Y - 2X + 1)\), [2]
  2. Var\((4Y - 5X + 3)\), [2]
  3. E\((X^2 Y)\). [3]

AnswerMarks Guidance
Answer/WorkingMark Guidance
\(E(4Y - 2X + 1) = 4E(Y) - 2E(X) + 1 = 4 \times 10 - 2 \times 17 + 1 = 7\)M1 A1 Use of
\(\text{Var}(4Y - 5X + 3) = 4^2\text{Var}(Y) + 5^2\text{Var}(X) = 16 \times 16 + 25 \times 64 = 1856\)M1 A1 cao
\(E(X^2) = \text{Var}(X) + (E(X))^2 = 64 + 17^2 = 353\)M1 Use of
\(E(X^2Y) = E(X^2) \times E(Y)\) \(E(X^2Y) = 353 \times 10 = 3530\)M1 A1 FT their \(E(X^2)\) provided \(\neq 17^2\); cao
Total [7]
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(4Y - 2X + 1) = 4E(Y) - 2E(X) + 1 = 4 \times 10 - 2 \times 17 + 1 = 7$ | M1 A1 | Use of |
| $\text{Var}(4Y - 5X + 3) = 4^2\text{Var}(Y) + 5^2\text{Var}(X) = 16 \times 16 + 25 \times 64 = 1856$ | M1 A1 | cao |
| $E(X^2) = \text{Var}(X) + (E(X))^2 = 64 + 17^2 = 353$ | M1 | Use of |
| $E(X^2Y) = E(X^2) \times E(Y)$ $E(X^2Y) = 353 \times 10 = 3530$ | M1 A1 | FT their $E(X^2)$ provided $\neq 17^2$; cao |
| **Total [7]** | | |
The random variable $X$ has mean 17 and variance 64. The independent random variable $Y$ has mean 10 and variance 16. Find the value of

\begin{enumerate}[label=(\alph*)]
\item E$(4Y - 2X + 1)$, [2]

\item Var$(4Y - 5X + 3)$, [2]

\item E$(X^2 Y)$. [3]
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 2 2023 Q1 [7]}}