Mathematics · Oct / Nov 2024 · Paper 61 · Question 6
YMA01
As printed in the paper
The question as text
6. A biased die with six faces is rolled. The discrete random variable X represents the score which is uppermost. The cumulative distribution function of X is shown in the table below. 1 2 3 4 5 6 x F(x) 0.1 0.2 3k 5k 7k 10k
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[1]
10
=
1
=
0.1
k
k
[
]
6(b)[3]
e.g. P(
X =
1)
=
0.1
and
P(
X =
2)
=
F(2)
−
F(1)
=
0.1
[
]
e.g.
P(
3)
F(3)
F(2)
0.1
X =
=
−
=
X
1
2
3
4
5
6
P(X = x)
0.1
0.1
0.1
0.2
0.2
0.3
= [
]
6(c)[3]
a
+
a
+
a
+
b
+
b
+
b
+
0.11 0.05
+
1
3
a
+
3
b
=
0.84
[
]
a
+
2
a
+
3
a
+
4
b
+
5
b
+
6
b
+
0.77
+
0.4
=
4.02
6
a
+
15
b
=
2.85
e.g. 9
=
1.35
=
0.15
*
a
a
6(d)[2]
b = 0.13
E(
2
)
=
1
2
×
0.15
+
2
2
×
0.15
+
3
2
×
0.15
+
4
2
×
'0.13' 5
+
2
×
'0.13' 6
+
2
×
'0.13'
Y
+
7
2
×
0.11 8
+
2
×
0.05
=
[20.7]*
[
]
6(e)[3]
Var( )
Y =
20.7
−
4.02 [
2
=
4.5396]
Var(5
−
2 )
Y
= 4Var( )
Y = 4 '4.5396'
×
= 18.1584 awrt 18.2
6(f)[2]
'0.1' 0.15
×
+
'0.1' 0.15
×
=
0.03
Notes
(a)
B1
For k = 0.1 oe
(b)
B1
For correct use of F(x) to find 2 probabilities May be implied by two correct probabilities
For correct use of F(x) to find one other probability. May be implied by one other correct
M1
probability
For a fully correct probability distribution. Need not be in a table but probabilities must be
A1
attached to the correct X values
For use of the sum of the probabilities = 1 to form a linear equation in a and b (May be implied
(c)
M1
by 3
a
+
3
b
=
0.84
)
For use of
×
P(
=
)
=
4.02
to form a linear equation in a and b
y
Y
y
M1
(May be implied by6
+
15
=
2.85
)
a
b
A1*
Answer is given so there must be a correct line between the 2 equations and the given answer
For finding E(Y2) with their b (At least 4 correct terms). Values for a and b must be substituted
(d)
M1
into their expression for E(Y2)
For a fully correct expression (no incorrect working seen)
A1*
eg. 14 0.15
×
+
77 0.13
×
+
49 0.11 64 0.05
×
+
×
Allow 0.15 + 0.6 + 1.35 + 2.08 + 3.25 + 4.68 + 5.39 + 3.2
For a correct expression for Var(Y). May be implied by awrt 4.54
(e)
M1
Allow E(5 – 2Y) =
−
76 25
oe and E((5 – 2Y)2) = 137 5
For use of 4Var(Y) ft their Var(Y) provided Var(Y) is not 20.7 or 4.02 Do not allow 5 + 4Var(Y)
M1
(
)
2
Allow Var(5 – 2Y) =
137
−−
76
ft their E(5 – 2Y) and their E((5 – 2Y)2)
5
25
A1
awrt 18.2
(f)
M1
For use of P(X = 1) × P(Y = 2) + P(X = 2) × P(Y = 1) ft their X distribution probabilities
A1
Cao
−
1
Red
n
2
n
+
1
n
n
n + and
+ in the correct places
2
n
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