CAIE S2 2019 November — Question 5 9 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2019
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeSingle sum threshold probability
DifficultyStandard +0.3 This question tests standard linear combinations of normal distributions with straightforward applications. Part (i) requires summing 10 independent normals (routine variance scaling), and part (ii) involves forming L - 2S and standardizing. Both are textbook exercises requiring only formula application and normal table lookup, making it slightly easier than average.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

5 The masses, in grams, of large boxes of chocolates and small boxes of chocolates have the distributions \(\mathrm { N } ( 325,6.1 )\) and \(\mathrm { N } ( 167,5.6 )\) respectively.
  1. Find the probability that the total mass of 10 randomly chosen large boxes of chocolates is less than 3240 g .
  2. Find the probability that the mass of a randomly chosen large box of chocolates is more than twice the mass of a randomly chosen small box of chocolates.

Question 5(i):
AnswerMarks Guidance
AnswerMark Guidance
mean \(= 3250\), var. \(= 61\)B1 Or mean \(= 325\), var. \(= \frac{6.1}{10}\)
\(\frac{3240 - 3250}{\sqrt{61}} = -1.280\)M1 Standardise with their values (no mixed methods)
\(\phi(-1.280) = 1 - \phi(1.280)\)M1 Area consistent with their figures
\(0.100\)A1 Allow 0.1
4
Question 5(ii):
AnswerMarks Guidance
AnswerMark Guidance
\(E(D) = 325 - 2 \times 167 = -9\)B1 Accept \(\pm 9\)
\(\text{Var}(D) = 6.1 + 2^2 \times 5.6 = 28.5\)B1
\(\frac{0-(-9)}{\sqrt{28.5}} = 1.686\)M1 Standardising with *their* values. Must have a combination attempt on denominator and \(\sqrt{\phantom{x}}\)
\(1 - \phi(1.686)\)M1 Area consistent with their figures
\(0.0459\)A1
5
## Question 5(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| mean $= 3250$, var. $= 61$ | B1 | Or mean $= 325$, var. $= \frac{6.1}{10}$ |
| $\frac{3240 - 3250}{\sqrt{61}} = -1.280$ | M1 | Standardise with their values (no mixed methods) |
| $\phi(-1.280) = 1 - \phi(1.280)$ | M1 | Area consistent with their figures |
| $0.100$ | A1 | Allow 0.1 |
| | **4** | |

## Question 5(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $E(D) = 325 - 2 \times 167 = -9$ | B1 | Accept $\pm 9$ |
| $\text{Var}(D) = 6.1 + 2^2 \times 5.6 = 28.5$ | B1 | |
| $\frac{0-(-9)}{\sqrt{28.5}} = 1.686$ | M1 | Standardising with *their* values. Must have a combination attempt on denominator and $\sqrt{\phantom{x}}$ |
| $1 - \phi(1.686)$ | M1 | Area consistent with their figures |
| $0.0459$ | A1 | |
| | **5** | |
5 The masses, in grams, of large boxes of chocolates and small boxes of chocolates have the distributions $\mathrm { N } ( 325,6.1 )$ and $\mathrm { N } ( 167,5.6 )$ respectively.\\
(i) Find the probability that the total mass of 10 randomly chosen large boxes of chocolates is less than 3240 g .\\

(ii) Find the probability that the mass of a randomly chosen large box of chocolates is more than twice the mass of a randomly chosen small box of chocolates.\\

\hfill \mbox{\textit{CAIE S2 2019 Q5 [9]}}