Further Mathematics · OCR H245 · Option S · Topic 5.04

Linear combinations of random variables

Linear combinations of any random variables, linear combinations of normal random variables.

2 sub-topics
1 in Year 12, 1 in Year 13
24 practice questions
2 animated lessons

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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

5.04aLinear combinations of any random variablesY13
5.04bLinear combinations of normal random variablesY12

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 24 questions in this topic, free, no account needed.

Linear combinations of random variables

Not needed for Year 12Easy

The mean of a linear combination · Year 13 · 1 mark

$\mathrm{E}(X) = 5$ and $\mathrm{E}(Y) = 5$ . What is $\mathrm{E}(3X - Y)$ ?
(1)

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$\mathrm{E}(aX + bY)$ ${}= a\mathrm{E}(X)$ ${}+ b\mathrm{E}(Y)$ , whether or not $X$ and $Y$ are independent. $C$ : 10

Total1

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Combinations of random variables

Not needed for Year 12Medium

The variance of a combination · Year 13 · 1 mark

$X$ and $Y$ are independent, with $\operatorname{Var}(X) = 5$ and $\operatorname{Var}(Y) = 1$ . What is $\operatorname{Var}(2X - 3Y)$ ?
(1)

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$\operatorname{Var}(aX + bY)$ ${}= a^2\operatorname{Var}(X)$ ${}+ b^2\operatorname{Var}(Y)$ for independent variables. Variances add, even for a difference. $B$ : 29

Total1

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Combinations of random variables

Not needed for Year 12Hard

Combinations of normal random variables · Year 13 · 5 marks

The masses of small parcels are A ~ N(15, 1.5²) grams and of large parcels B ~ N(40, 2.5²) grams, all independent.

Find the probability that the total mass of three small parcels is greater than the mass of one large parcel.
(5)

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Define the difference and find $\mathrm{P}(D > 0)$ . $D = A_1 + A_2 + A_3 - B$
$\mathrm{E}(D) = 5$
Variances add, even for a difference. $\operatorname{Var}(D)$ ${}= 3(1.5^2) + 2.5^2$ ${}= 13$
$\mathrm{P}(D > 0) = 0.9172$

Total5

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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What the specification says

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