Question 5 - A-Level Maths

Edexcel S1 2023 June — Question 5 13 marks

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2023
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Discrete Probability Distributions
Type	Piecewise or conditional probability function
Difficulty	Standard +0.3 This is a straightforward S1 probability distribution question requiring standard techniques: summing probabilities to find k, calculating probabilities over intervals, finding E(Y), solving an inequality Y ≥ 15-2Y, and applying the linear transformation formula Var(aY+b) = a²Var(Y). All steps are routine applications of formulas with no novel insight required, making it slightly easier than average.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance

A discrete random variable $Y$ has probability function $$\mathrm{P}(Y = y) = \begin{cases} k(3 - y) & y = 1, 2 \\ k(y^2 - 8) & y = 3, 4, 5 \\ k & y = 6 \\ 0 & \text{otherwise} \end{cases}$$ where $k$ is a constant.

Show that $k = \frac{1}{30}$ [2]

Find the exact value of

P$(1 < Y \leqslant 4)$ [2]
E$(Y)$ [2]

The random variable $X = 15 - 2Y$

Calculate P$(Y \geqslant X)$ [3]
Calculate Var$(X)$ [4]

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5(a)	P(Y = y) 2k k k 8k 17k k	M1

2k+k+k+8k+17k+k=1 or 30k =1 k = 1 *

Answer	Marks
 30	A1*

(2)

Answer	Marks	Guidance
(b)	k+k+8k or 1−(2k+17k+k)	M1

=1 oe awrt 0.333

Answer	Marks
3	A1

(2)

Answer	Marks	Guidance
(c)	(12k)+(2k)+(3k)+(48k)+(517k)+(6k) =	M1

13 oe awrt 4.33

Answer	Marks
3	A1

(2)

Answer	Marks	Guidance
(d)	P(Y 15−2Y) or [X =] 13 11 9 7 5 3 only or [Y =] 5 or 6 only	M1

"17" "1"

[P(Y 5) = P(Y =5)+P(Y =6) ] = +

Answer	Marks
30 30	M1

3 = oe

Answer	Marks
5	A1ft

(3)

Answer	Marks	Guidance
(e)	Var (X) = 4Var(Y)	M1

[E(Y2)=](12k)+(22k)+(32k)+(428k)+(5217k)+(62k)=302or awrt 20.1

Answer	Marks
 15 	M1

[Var(Y)=]" 302 "− ( " 13 " )2 = 61 or awrt 1.36

Answer	Marks
15 3  45 	M1d

 Var (X)= 244 oe awrt 5.42

Answer	Marks
45	A1
ALT for 1st 3 marks	(4)
[	E(X) =] (132k)+(11k)+(9k)+(78k)+(517k)+(3k)=19or awrt 6.33
 3 	M1
[	E(X2)=](1322k)+(112k)+(92k)+(728k)+(5217k)+(32k)= 683or awrt 45.5
 15 	M1
[	Var(X)=]"683"− ( "190" )2
15 30	M1d
Notes	Total 13
(a)	M1 for finding the probabilities in terms of k. The individual probabilities must be seen either in a table

or in the calculation (but do not need to be simplified)

Answer	Marks
A1* Method mark must be awarded. For a correct equation which would lead to k =1	30*

NB Verification - 2 ( 1 ) +( 1 )+( 1 )+8 ( 1 )+17 ( 1 )+( 1 )=1 gains M1 A0

30 30 30 30 30 30

Answer	Marks
(b)	M1 for using P(Y =2)+P(Y =3)+P(Y =4)or 1−P(Y =1)+P(Y =5)+P(Y =6)Allow in terms of k or

with k =130subst or with their probabilities. Do not allow in terms of y

A1 awrt 0.333

Answer	Marks
(c)	M1 for using xP(x) At least 3 terms given Allow with k =130subst or ft their probabilities.

A1 awrt 4.33

Answer	Marks
(d)	M1 forming correct inequality in Y or 13,11,9,7,5,3 seen anywhere or for 5 and 6 only. Implied by 2nd M1

M1 findingtheir P(Y =5)+their P(Y =6) or P(X =5)+P(X =3) eg 17k + k

A1ft ft their probabilities

Answer	Marks
(e)	M1 written or used 4Var(Y) (may come at the end of the calculation) or written or used E(X) allow awrt

6.33 NB condone −22Var(Y)if used 4Var(Y)

M1 Correct method, at least 3 products correct, to find E(Y2)or E(X2)condone incorrect labels

M1d dep on the 2nd M mark being awarded. For correct use of E(Y2)−[E(Y)]2 or E(X2)−[E(X)]2 For

the ALT In addition to the 2nd M1 the 1st M1 must be awarded. Condone incorrect labelling

A1 awrt 5.42

Answer	Marks	Guidance
P(Y = y)	2k	k
Qu	Scheme	Marks

Question 5:
--- 5(a) ---
5(a) | P(Y = y) 2k k k 8k 17k k | M1
2k+k+k+8k+17k+k=1 or 30k =1 k = 1 *
 30 | A1*
(2)
(b) | k+k+8k or 1−(2k+17k+k) | M1
=1 oe awrt 0.333
3 | A1
(2)
(c) | (12k)+(2k)+(3k)+(48k)+(517k)+(6k) = | M1
13 oe awrt 4.33
3 | A1
(2)
(d) | P(Y 15−2Y) or [X =] 13 11 9 7 5 3 only or [Y =] 5 or 6 only | M1
"17" "1"
[P(Y 5) = P(Y =5)+P(Y =6) ] = +
30 30 | M1
3
= oe
5 | A1ft
(3)
(e) | Var (X) = 4Var(Y) | M1
[E(Y2)=](12k)+(22k)+(32k)+(428k)+(5217k)+(62k)=302or awrt 20.1
 15  | M1
[Var(Y)=]" 302 "− ( " 13 " )2 = 61 or awrt 1.36
15 3  45  | M1d
 Var (X)= 244 oe awrt 5.42
45 | A1
ALT for 1st 3 marks | (4)
[ | E(X) =] (132k)+(11k)+(9k)+(78k)+(517k)+(3k)=19or awrt 6.33
 3  | M1
[ | E(X2)=](1322k)+(112k)+(92k)+(728k)+(5217k)+(32k)= 683or awrt 45.5
 15  | M1
[ | Var(X)=]"683"− ( "190" )2
15 30 | M1d
Notes | Total 13
(a) | M1 for finding the probabilities in terms of k. The individual probabilities must be seen either in a table
or in the calculation (but do not need to be simplified)
A1* Method mark must be awarded. For a correct equation which would lead to k =1 | 30*
NB Verification - 2 ( 1 ) +( 1 )+( 1 )+8 ( 1 )+17 ( 1 )+( 1 )=1 gains M1 A0
30 30 30 30 30 30
(b) | M1 for using P(Y =2)+P(Y =3)+P(Y =4)or 1−P(Y =1)+P(Y =5)+P(Y =6)Allow in terms of k or
with k =130subst or with their probabilities. Do not allow in terms of y
A1 awrt 0.333
(c) | M1 for using xP(x) At least 3 terms given Allow with k =130subst or ft their probabilities.
A1 awrt 4.33
(d) | M1 forming correct inequality in Y or 13,11,9,7,5,3 seen anywhere or for 5 and 6 only. Implied by 2nd M1
M1 findingtheir P(Y =5)+their P(Y =6) or P(X =5)+P(X =3) eg 17k + k
A1ft ft their probabilities
(e) | M1 written or used 4Var(Y) (may come at the end of the calculation) or written or used E(X) allow awrt
6.33 NB condone −22Var(Y)if used 4Var(Y)
M1 Correct method, at least 3 products correct, to find E(Y2)or E(X2)condone incorrect labels
M1d dep on the 2nd M mark being awarded. For correct use of E(Y2)−[E(Y)]2 or E(X2)−[E(X)]2 For
the ALT In addition to the 2nd M1 the 1st M1 must be awarded. Condone incorrect labelling
A1 awrt 5.42
P(Y = y) | 2k | k | k | 8k | 17k | k
Qu | Scheme | Marks

Show LaTeX source

A discrete random variable $Y$ has probability function

$$\mathrm{P}(Y = y) = \begin{cases}
k(3 - y) & y = 1, 2 \\
k(y^2 - 8) & y = 3, 4, 5 \\
k & y = 6 \\
0 & \text{otherwise}
\end{cases}$$

where $k$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{1}{30}$ [2]
\end{enumerate}

Find the exact value of

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item P$(1 < Y \leqslant 4)$ [2]
\item E$(Y)$ [2]
\end{enumerate}

The random variable $X = 15 - 2Y$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Calculate P$(Y \geqslant X)$ [3]
\item Calculate Var$(X)$ [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2023 Q5 [13]}}

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