Question 4 - A-Level Maths

CAIE S2 2023 June — Question 4 6 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2023
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear combinations of normal random variables
Type	Multiple period profit/loss calculation
Difficulty	Standard +0.3 This is a straightforward application of linear combinations of normal random variables. Students need to recognize that total profit = 50 × (sum of 7 independent normal variables), apply the scaling and summation properties (mean multiplies by 350, variance by 50² × 7), then perform a standard normal probability calculation. While it requires multiple steps, each is routine for S2 level with no novel insight needed.
Spec	5.04a Linear combinations: E(aX+bY), Var(aX+bY)

4 The mass, in tonnes, of steel produced per day at a factory is normally distributed with mean 65.2 and standard deviation 3.6. It can be assumed that the mass of steel produced each day is independent of other days. The factory makes $\$ 50$ profit on each tonne of steel produced. Find the probability that the total profit made in a randomly chosen 7-day week is less than $\$ 22000$.

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Question 4:

Method 1: Based on mass

Answer	Marks	Guidance
Answer	Mark	Guidance
Mean $= 7 \times 65.2 = 456.4$	B1
$\text{Var} = 7 \times 3.6^2\ [= 90.72]$	M1
$22000/50 = 440$ used in standardising equation	M1
$\frac{440 - 456.4}{\sqrt{90.72}}\ [= -1.722]$ no mixed methods	M1	For standardising with their values. No mixed methods.
$\Phi(-1.722) = 1 - \Phi(1.722)$	M1	For correct probability area consistent with their values.
$= 0.0425$ or $0.0426$	A1	Note: accept alt method using per day. $N(65.2, \frac{3.6^2}{7})$. No mixed methods.

Method 2: Based on profit

Answer	Marks	Guidance
Answer	Mark	Guidance
Mean $= 7 \times 65.2 \times 50 = 22820$	B1
$\text{Var} = 7 \times 3.6^2$	M1
$\text{Var} = 50^2 \times 90.72\ [= 226800]$	M1
$\frac{22000 - 22820}{\sqrt{226800}}\ [= -1.722]$ no mixed methods	M1	For standardising with their values. No mixed methods.
$\Phi(-1.722) = 1 - \Phi(1.722)$	M1	For correct probability area consistent with their values.
$= 0.0425$ or $0.0426$	A1
6

## Question 4:

**Method 1: Based on mass**

| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 7 \times 65.2 = 456.4$ | B1 | |
| $\text{Var} = 7 \times 3.6^2\ [= 90.72]$ | M1 | |
| $22000/50 = 440$ used in standardising equation | M1 | |
| $\frac{440 - 456.4}{\sqrt{90.72}}\ [= -1.722]$ no mixed methods | M1 | For standardising with their values. No mixed methods. |
| $\Phi(-1.722) = 1 - \Phi(1.722)$ | M1 | For correct probability area consistent with their values. |
| $= 0.0425$ or $0.0426$ | A1 | Note: accept alt method using per day. $N(65.2, \frac{3.6^2}{7})$. No mixed methods. |

**Method 2: Based on profit**

| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 7 \times 65.2 \times 50 = 22820$ | B1 | |
| $\text{Var} = 7 \times 3.6^2$ | M1 | |
| $\text{Var} = 50^2 \times 90.72\ [= 226800]$ | M1 | |
| $\frac{22000 - 22820}{\sqrt{226800}}\ [= -1.722]$ no mixed methods | M1 | For standardising with their values. No mixed methods. |
| $\Phi(-1.722) = 1 - \Phi(1.722)$ | M1 | For correct probability area consistent with their values. |
| $= 0.0425$ or $0.0426$ | A1 | |
| | **6** | |

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4 The mass, in tonnes, of steel produced per day at a factory is normally distributed with mean 65.2 and standard deviation 3.6. It can be assumed that the mass of steel produced each day is independent of other days. The factory makes $\$ 50$ profit on each tonne of steel produced.

Find the probability that the total profit made in a randomly chosen 7-day week is less than $\$ 22000$.\\

\hfill \mbox{\textit{CAIE S2 2023 Q4 [6]}}

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