WJEC Further Unit 5 2022 June — Question 2 15 marks

Exam BoardWJEC
ModuleFurther Unit 5 (Further Unit 5)
Year2022
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeSum versus sum comparison
DifficultyChallenging +1.2 This is a standard Further Maths statistics question on linear combinations of normal variables. Parts (a) and (b) are routine applications of normal distribution and sum properties. Part (c) requires setting up the inequality 5X_B ≥ 2(3X_A), which becomes a comparison of two normal distributions—a step beyond basic but still a well-practiced technique in Further Maths. The multi-part structure and the final part requiring algebraic manipulation of the inequality elevate it slightly above average difficulty.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

2. Geraint is a beekeeper. The amounts of honey, \(X \mathrm {~kg}\), that he collects annually, from each hive are modelled by the normal distribution \(\mathrm { N } \left( 15,5 ^ { 2 } \right)\). At location \(A\), Geraint has three hives and at location \(B\) he has five hives. You may assume that the amounts of honey collected from the eight hives are independent of each other.
    1. Find the probability that Geraint collects more than 14 kg of honey from the first hive at location \(A\).
    2. Find the probability that he collects more than 14 kg of honey from exactly two out of the three hives at location \(A\).
  1. Find the probability that the total amount of honey that Geraint collects from all eight hives is more than 160 kg .
  2. Find the probability that Geraint collects at least twice as much honey from location B as from location A.

Question 2:
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(X > 14) = 0.5793\)M1A1 M1 for correct method (calculator or standardizing); A1 FT for M1A1 "their (i)" and "1-(i)" awrt 0.423 or 0.424
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(X > 14 \text{ for two out of three}) = 0.5793^2 \times 0.4207 \times 3\)M1
\(= 0.4235\)A1
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Let \(T = X_1 + X_2 + X_3 + \cdots + X_8\)
\(E(T) = 120\)B1
\(\text{Var}(T) = 8\text{Var}(X)\)M1
\(\text{Var}(T) = 200\)A1
\(P(T > 160) = 0.00234\) (3sf)A1 cao; 0.00233 from tables
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Let \(A = X_1 + X_2 + X_3\); Let \(B = X_1 + X_2 + X_3 + X_4 + X_5\); \(A \sim N(45, 75)\) and \(B \sim N(75, 125)\)B1 si
Consider \(U = B - 2A\); \(E(U) = -15\)M1, A1 M1A0 for \(E(U) = 105\) from \(U = 2B - A\)
\(\text{Var}(U) = \text{Var}(B) + 2^2\text{Var}(A) = 425\)M1, A1 M1A1 for \(\text{Var}(U) = 575\) from \(U = 2B - A\)
\(P(U > 0)\)m1 Dependent on 1st M1 and \(U = B - 2A\); m0A0 if \(U = 2B - A\)
\(= 0.2334\)A1 cao; 0.23270 from tables
# Question 2:

## Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 14) = 0.5793$ | M1A1 | M1 for correct method (calculator or standardizing); A1 FT for M1A1 "their (i)" and "1-(i)" awrt 0.423 or 0.424 |

## Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 14 \text{ for two out of three}) = 0.5793^2 \times 0.4207 \times 3$ | M1 | |
| $= 0.4235$ | A1 | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $T = X_1 + X_2 + X_3 + \cdots + X_8$ | | |
| $E(T) = 120$ | B1 | |
| $\text{Var}(T) = 8\text{Var}(X)$ | M1 | |
| $\text{Var}(T) = 200$ | A1 | |
| $P(T > 160) = 0.00234$ (3sf) | A1 | cao; 0.00233 from tables |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $A = X_1 + X_2 + X_3$; Let $B = X_1 + X_2 + X_3 + X_4 + X_5$; $A \sim N(45, 75)$ and $B \sim N(75, 125)$ | B1 | si |
| Consider $U = B - 2A$; $E(U) = -15$ | M1, A1 | M1A0 for $E(U) = 105$ from $U = 2B - A$ |
| $\text{Var}(U) = \text{Var}(B) + 2^2\text{Var}(A) = 425$ | M1, A1 | M1A1 for $\text{Var}(U) = 575$ from $U = 2B - A$ |
| $P(U > 0)$ | m1 | Dependent on 1st M1 and $U = B - 2A$; m0A0 if $U = 2B - A$ |
| $= 0.2334$ | A1 | cao; 0.23270 from tables |

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2. Geraint is a beekeeper. The amounts of honey, $X \mathrm {~kg}$, that he collects annually, from each hive are modelled by the normal distribution $\mathrm { N } \left( 15,5 ^ { 2 } \right)$. At location $A$, Geraint has three hives and at location $B$ he has five hives. You may assume that the amounts of honey collected from the eight hives are independent of each other.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that Geraint collects more than 14 kg of honey from the first hive at location $A$.
\item Find the probability that he collects more than 14 kg of honey from exactly two out of the three hives at location $A$.
\end{enumerate}\item Find the probability that the total amount of honey that Geraint collects from all eight hives is more than 160 kg .
\item Find the probability that Geraint collects at least twice as much honey from location B as from location A.
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 5 2022 Q2 [15]}}