CAIE S2 2017 November — Question 6 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2017
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeExpectation and variance with context application
DifficultyModerate -0.3 This is a straightforward application of linear transformations of normal variables and combining independent normals. Part (i) requires only E(aX) = aE(X) and Var(aX) = a²Var(X), while part (ii) needs recognizing that aX + bY is normal with combined mean/variance, then a standard normal probability calculation. All steps are routine S2 techniques with no problem-solving insight required, making it slightly easier than average.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

6 The numbers of barrels of oil, in millions, extracted per day in two oil fields \(A\) and \(B\) are modelled by the independent random variables \(X\) and \(Y\) respectively, where \(X \sim \mathrm {~N} \left( 3.2,0.4 ^ { 2 } \right)\) and \(Y \sim \mathrm {~N} \left( 4.3,0.6 ^ { 2 } \right)\). The income generated by the oil from the two fields is \(\\) 90\( per barrel for \)A\( and \)\\( 95\) per barrel for \(B\).
  1. Find the mean and variance of the daily income, in millions of dollars, generated by field \(A\). [3]
  2. Find the probability that the total income produced by the two fields in a day is at least \(\\) 670$ million.

Question 6(i):
AnswerMarks Guidance
AnswerMark Guidance
Mean \(= 3.2 \times 90 = 288\)B1
Variance \(= 0.4^2 \times 90^2\)M1
\(= 1296\)A1
Total: 3
Question 6(ii):
AnswerMarks Guidance
AnswerMark Guidance
Mean \(= \text{'288'} + 4.3 \times 95 = 696.5\)B1 FT
Variance \(= \text{'1296'} + 0.6^2 \times 95^2 = 4545\)B1 FT FT their (i)
\(\frac{670-696.5}{\sqrt{4545}}\) \((= -0.393)\)M1 FT Var provided both given Vars used standardising (ignore cc); no sd/Var mix
\(1 - \phi(\text{'-0.393'}) = \phi(\text{'0.393'})\)M1 correct area consistent with their working (i.e. their mean)
\(= 0.653\) (3 sf)A1
Total: 5
## Question 6(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 3.2 \times 90 = 288$ | B1 | |
| Variance $= 0.4^2 \times 90^2$ | M1 | |
| $= 1296$ | A1 | |
| **Total: 3** | | |

## Question 6(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \text{'288'} + 4.3 \times 95 = 696.5$ | B1 FT | |
| Variance $= \text{'1296'} + 0.6^2 \times 95^2 = 4545$ | B1 FT | FT their (i) |
| $\frac{670-696.5}{\sqrt{4545}}$ $(= -0.393)$ | M1 | FT Var provided both given Vars used standardising (ignore cc); no sd/Var mix |
| $1 - \phi(\text{'-0.393'}) = \phi(\text{'0.393'})$ | M1 | correct area consistent with their working (i.e. their mean) |
| $= 0.653$ (3 sf) | A1 | |
| **Total: 5** | | |

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6 The numbers of barrels of oil, in millions, extracted per day in two oil fields $A$ and $B$ are modelled by the independent random variables $X$ and $Y$ respectively, where $X \sim \mathrm {~N} \left( 3.2,0.4 ^ { 2 } \right)$ and $Y \sim \mathrm {~N} \left( 4.3,0.6 ^ { 2 } \right)$. The income generated by the oil from the two fields is $\$ 90$ per barrel for $A$ and $\$ 95$ per barrel for $B$.\\
(i) Find the mean and variance of the daily income, in millions of dollars, generated by field $A$. [3]\\

(ii) Find the probability that the total income produced by the two fields in a day is at least $\$ 670$ million.\\

\hfill \mbox{\textit{CAIE S2 2017 Q6 [8]}}