Mathematics · Oct / Nov 2022 · Paper 61 · Question 7

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The question as text

7. Adana selects one number at random from the distribution of X which has the following probability distribution. 0 5 10 x P(X = x) 0.1 0.2 0.7

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Show the answer and mark scheme

7(a)[1]
0.8
7(b)[1]
[0
2
×
0.1 ]5
+
2
×
0.2 10
+
2
×
0.7
=
75*
7(c)[3]
E(
X =
)
[0 0.1 ]5 0.2 10 0.7[ 8]
×
+
×
+
×
=
2
Var(
X =
)
75 ('8')
−
X =11
Var(
)
2
2
7(d)[2]
Var(4
−
3
X
)
=
3
×
Var(
X
)[ 3
=
×
"11"]
= 99
(e) P((0, 5), (0, 10), (5, 10))
P((0, 0), (5, 5), (10, 10)
=0.1 0.2
×
+
0.1 0.7
×
+
0.2 0.7
×
= 1 – (0.12 + 0.22 + 0.72)
[
]
(
(
)
)
=
0.02
+
0.07
+
0.14
=
0.5 1– 0.01 0.04
+
+
0.49
= 0.23
= 0.23
(f) Products: 0, 25, 50, 100
P(
25)
0.2
2
P(
D =
0)
=
0.1 0.1 0.1 0.1[ 0.19]
+
−
×
=
D =
=
P(
D =
100)
=
0.7
2
P(
D =
50)
=
2 0.2 0.7
×
×
0
25
50
100
D
P(D = d)
0.19
0.04
0.28
0.49
Notes
(a) B1 1
8 oe Allow 0.125 Do not ISW
(b) B1*cso correct expression and 75 with no errors seen. Allow
5
2
×
0.2 10
+
2
×
0.7
(c) M1 correct method to find mean. If no method seen award if 8 is seen
M1 attempt at expression for variance ie 75 –( their E(X))2
A1 11 cao
(d) M1 Use of
( 3)
−
2
×
Var(
)
X
(condone 3 rather than –3 and missing bracket if final answer is
(
)
(
)
(
)
correct) or
4
2
×
0.1
+ −
11
2
×
0.2
+ −
26
2
×
0.7
−−
20
2
condone 11, 26 and 20
A1ft 99 or ft 9 × ‘their (c)’
(e) 1st M1 for at least one correct product. NB may be combined eg 0.3 0.7
×
but not in the
numerator or denominator of a fraction
2nd M1 a fully correct expression oe e.g. 0.1 0.2
×
+
0.3 0.7
×
A1 0.23 oe
(f) B1 all 4 correct products with no incorrect extras unless they have a probability of 0
associated with them
1st M1 A correct method to find 1 of the 4 probabilities. Does not need to be associated with
the correct product
ALT P(
D =
0)
=
0.1 0.1 2 0.1 0.2
×
+ ×
×
+ ×
2 0.1 0.7[ 0.19]
×
=
2nd M1 A correct method to find 3 of the 4 probabilities or 2 of the 4 probabilities if the
total of the 4 probabilities is 1 Must be associated with the correct product
A1 all four correct probabilities (oe) associated with the correct products
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
7(e)[3]
0.8
[0
2
×
0.1 ]5
+
2
×
0.2 10
+
2
×
0.7
=
75*
E(
X =
)
[0 0.1 ]5 0.2 10 0.7[ 8]
×
+
×
+
×
=
2
Var(
X =
)
75 ('8')
−
X =11
Var(
)
2
2
Var(4
−
3
X
)
=
3
×
Var(
X
)[ 3
=
×
"11"]
= 99
(e) P((0, 5), (0, 10), (5, 10))
P((0, 0), (5, 5), (10, 10)
=0.1 0.2
×
+
0.1 0.7
×
+
0.2 0.7
×
= 1 – (0.12 + 0.22 + 0.72)
[
]
(
(
)
)
=
0.02
+
0.07
+
0.14
=
0.5 1– 0.01 0.04
+
+
0.49
= 0.23
= 0.23
(f) Products: 0, 25, 50, 100
P(
25)
0.2
2
P(
D =
0)
=
0.1 0.1 0.1 0.1[ 0.19]
+
−
×
=
D =
=
P(
D =
100)
=
0.7
2
P(
D =
50)
=
2 0.2 0.7
×
×
0
25
50
100
D
P(D = d)
0.19
0.04
0.28
0.49
Notes
(a) B1 1
8 oe Allow 0.125 Do not ISW
(b) B1*cso correct expression and 75 with no errors seen. Allow
5
2
×
0.2 10
+
2
×
0.7
(c) M1 correct method to find mean. If no method seen award if 8 is seen
M1 attempt at expression for variance ie 75 –( their E(X))2
A1 11 cao
(d) M1 Use of
( 3)
−
2
×
Var(
)
X
(condone 3 rather than –3 and missing bracket if final answer is
(
)
(
)
(
)
correct) or
4
2
×
0.1
+ −
11
2
×
0.2
+ −
26
2
×
0.7
−−
20
2
condone 11, 26 and 20
A1ft 99 or ft 9 × ‘their (c)’
(e) 1st M1 for at least one correct product. NB may be combined eg 0.3 0.7
×
but not in the
numerator or denominator of a fraction
2nd M1 a fully correct expression oe e.g. 0.1 0.2
×
+
0.3 0.7
×
A1 0.23 oe
(f) B1 all 4 correct products with no incorrect extras unless they have a probability of 0
associated with them
1st M1 A correct method to find 1 of the 4 probabilities. Does not need to be associated with
the correct product
ALT P(
D =
0)
=
0.1 0.1 2 0.1 0.2
×
+ ×
×
+ ×
2 0.1 0.7[ 0.19]
×
=
2nd M1 A correct method to find 3 of the 4 probabilities or 2 of the 4 probabilities if the
total of the 4 probabilities is 1 Must be associated with the correct product
A1 all four correct probabilities (oe) associated with the correct products
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
7(f)[4]
0.8
[0
2
×
0.1 ]5
+
2
×
0.2 10
+
2
×
0.7
=
75*
E(
X =
)
[0 0.1 ]5 0.2 10 0.7[ 8]
×
+
×
+
×
=
2
Var(
X =
)
75 ('8')
−
X =11
Var(
)
2
2
Var(4
−
3
X
)
=
3
×
Var(
X
)[ 3
=
×
"11"]
= 99
(e) P((0, 5), (0, 10), (5, 10))
P((0, 0), (5, 5), (10, 10)
=0.1 0.2
×
+
0.1 0.7
×
+
0.2 0.7
×
= 1 – (0.12 + 0.22 + 0.72)
[
]
(
(
)
)
=
0.02
+
0.07
+
0.14
=
0.5 1– 0.01 0.04
+
+
0.49
= 0.23
= 0.23
(f) Products: 0, 25, 50, 100
P(
25)
0.2
2
P(
D =
0)
=
0.1 0.1 0.1 0.1[ 0.19]
+
−
×
=
D =
=
P(
D =
100)
=
0.7
2
P(
D =
50)
=
2 0.2 0.7
×
×
0
25
50
100
D
P(D = d)
0.19
0.04
0.28
0.49
Notes
(a) B1 1
8 oe Allow 0.125 Do not ISW
(b) B1*cso correct expression and 75 with no errors seen. Allow
5
2
×
0.2 10
+
2
×
0.7
(c) M1 correct method to find mean. If no method seen award if 8 is seen
M1 attempt at expression for variance ie 75 –( their E(X))2
A1 11 cao
(d) M1 Use of
( 3)
−
2
×
Var(
)
X
(condone 3 rather than –3 and missing bracket if final answer is
(
)
(
)
(
)
correct) or
4
2
×
0.1
+ −
11
2
×
0.2
+ −
26
2
×
0.7
−−
20
2
condone 11, 26 and 20
A1ft 99 or ft 9 × ‘their (c)’
(e) 1st M1 for at least one correct product. NB may be combined eg 0.3 0.7
×
but not in the
numerator or denominator of a fraction
2nd M1 a fully correct expression oe e.g. 0.1 0.2
×
+
0.3 0.7
×
A1 0.23 oe
(f) B1 all 4 correct products with no incorrect extras unless they have a probability of 0
associated with them
1st M1 A correct method to find 1 of the 4 probabilities. Does not need to be associated with
the correct product
ALT P(
D =
0)
=
0.1 0.1 2 0.1 0.2
×
+ ×
×
+ ×
2 0.1 0.7[ 0.19]
×
=
2nd M1 A correct method to find 3 of the 4 probabilities or 2 of the 4 probabilities if the
total of the 4 probabilities is 1 Must be associated with the correct product
A1 all four correct probabilities (oe) associated with the correct products
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom

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