Mathematics · Oct / Nov 2023 · Paper 61 · Question 4
YMA01
As printed in the paper
The question as text
4. The discrete random variable X has the following probability distribution. 1 2 3 4 x 1 1 3 2 P(X = x) 10 5 10 5 1 2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[1]
E
1 10
+
×
+
×
+
×
=
X
2
5
3
10
4
5
5
1
2
1
1
2
1
1
2
3
1
2
2
5
4(b)[3]
E
=
1
2
×
+
×
+
×
+
×
=
X
10
2
5
3
10
4
5
24
2
1
5
2
29
Var
=
−
=
24
5
600
X
[
]
(c) (i)
E( )
Y =
12
1
87
375
87
[
]
or If y : 30 15 10 7.5 then [
]
(ii)
Var( )
Y
=
30 'Var
2
'
=
Var( )
Y =
−
12
2
=
2
2
2
X
]30
[
4(c)[3]
E
1 10
+
×
+
×
+
×
=
X
2
5
3
10
4
5
5
1
2
1
1
2
1
1
2
3
1
2
2
5
E
=
1
2
×
+
×
+
×
+
×
=
X
10
2
5
3
10
4
5
24
2
1
5
2
29
Var
=
−
=
24
5
600
X
[
]
(c) (i)
E( )
Y =
12
1
87
375
87
[
]
or If y : 30 15 10 7.5 then [
]
(ii)
Var( )
Y
=
30 'Var
2
'
=
Var( )
Y =
−
12
2
=
2
2
2
X
]30
[
Y
<
20
<
20
X
>
1.5
or y : 30 15 10 7.5
X
9
P(
<
20)
=
P(
>
1.5)
=
Y
X
10
1
1
P(
X
=
2)
2
P(
Y
=
15)
5
2
dM1 A1
[
]
5
or [
]
P(
<
3|
<
20)
=
=
=
P(
<
3|
<
20)
=
=
=
X
Y
X
Y
P(
>
1.5)
9
9
P(
Y
<
20)
9
9
X
'
'
'
'
10
10
Notes
Value given, so must see sight of a correct expression, with no incorrect working seen. (Allow
(a)
B1*
equivalent expressions.)
1
2
For attempt at an expression for
with at least 3 correct terms
E
(b)
M1
X
(Allow equivalent expressions.) May be embedded in a correct expression for Var (X)
1
For a correct expression for
(Need not be simplified) ft a stated value of
Var
E
M1
X
A1
Cao Allow awrt 0.0483
(c) (i)
For [E(Y)] = 12
B1
1
1
375
(ii)
For correct use of
30 Var
2
ft their
Var
or
2 −
12
2
(May be implied by 87
2 oe)
M1
X
X
]87
For [
oe
Var( )
Y =
A1
2
Y <
20
(d)
M1
For a correct inequality for
or all 4 values of Y found (these may be seen in part (c))
9
For
(May be seen as the denominator (e.g 0.2 + 0.3 + 0.4 oe) in a ratio of
P(
Y <
20)
=
A1
10
probabilities and scores M1A1)
or P(
=
15)
Dependant on 1st M1 For P(
X
=
2)
Y
Allow P(1.5
<
X
<
3)
dM1
P(
X
>
1.5)
P(
Y
<
20)
P(
X
>
1.5)
or a correct ratio of probabilities ftP(
Y <
20)
For a correct numerator
A1
For 2
oe (Allow a decimal answer that is 3sf or better e.g. 0.222)
A1
9
4(d)[5]
Y
<
20
<
20
X
>
1.5
or y : 30 15 10 7.5
X
9
P(
<
20)
=
P(
>
1.5)
=
Y
X
10
1
1
P(
X
=
2)
2
P(
Y
=
15)
5
2
dM1 A1
[
]
5
or [
]
P(
<
3|
<
20)
=
=
=
P(
<
3|
<
20)
=
=
=
X
Y
X
Y
P(
>
1.5)
9
9
P(
Y
<
20)
9
9
X
'
'
'
'
10
10
Notes
Value given, so must see sight of a correct expression, with no incorrect working seen. (Allow
(a)
B1*
equivalent expressions.)
1
2
For attempt at an expression for
with at least 3 correct terms
E
(b)
M1
X
(Allow equivalent expressions.) May be embedded in a correct expression for Var (X)
1
For a correct expression for
(Need not be simplified) ft a stated value of
Var
E
M1
X
A1
Cao Allow awrt 0.0483
(c) (i)
For [E(Y)] = 12
B1
1
1
375
(ii)
For correct use of
30 Var
2
ft their
Var
or
2 −
12
2
(May be implied by 87
2 oe)
M1
X
X
]87
For [
oe
Var( )
Y =
A1
2
Y <
20
(d)
M1
For a correct inequality for
or all 4 values of Y found (these may be seen in part (c))
9
For
(May be seen as the denominator (e.g 0.2 + 0.3 + 0.4 oe) in a ratio of
P(
Y <
20)
=
A1
10
probabilities and scores M1A1)
or P(
=
15)
Dependant on 1st M1 For P(
X
=
2)
Y
Allow P(1.5
<
X
<
3)
dM1
P(
X
>
1.5)
P(
Y
<
20)
P(
X
>
1.5)
or a correct ratio of probabilities ftP(
Y <
20)
For a correct numerator
A1
For 2
oe (Allow a decimal answer that is 3sf or better e.g. 0.222)
A1
9
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