Mathematics · May / June 2022 · Paper 61 · Question 5
YMA01
As printed in the paper
The question as text
5. A red spinner is designed so that the score R is given by the following probability distribution. 2 3 4 5 6 r P(R = r) 0.25 0.3 0.15 0.1 0.2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[1]
R
=
×
+
×
+
×
+
×
+
×
(= 15.8*)
( )
5(b)[2]
s
d
R =
1
5
.8
−
3
.7
2
=
2.11
Standard deviation = 1.4525….
awrt 1.45 A1
5(c)[1]
d =
1
B1
0.1 0.2
0.1
1
5(d)[5]
+
+
+
a
+
b
= oe
M1
2 0.1 3 0.2
×
+ ×
+ ×
4 0.1 5
+
+
6
=
4.55
a
b
oe
M1
(
)
(
)
5 0.6
−
b
+
6
b
=
3.35
or
5
a
+
6 0.6
−
a
=
3.35
a
=
0.25
or
b =
0.35
c =
0.4
+
"0.25"
or
c = −
1 "0.35"
M1
c =
0.65
oe
A1
5(e)[2]
0.9 0.75 0.1
×
×
M1
= 0.0675
A1
5(f)[3]
For identifying that if Jessie scores 2, Pabel has no spin oe may be implied
[0.10 0 ]0.2 0.3
×+
×
+
0.1 0.15 "0.25" 0.1 "0.35" 0.2
×
+
×
+
×
M1
= 0.17
A1
Notes
(a)
B1
Correct calculation with all products seen (allow 1 + 2.7 + 2.4 + 2.5 + 7.2)
Figures may be seen in table before part (a). Condone missing addition signs if products seen in table.
(b)
Use of formula including the square root
M1
A1
awrt 1.45 (correct answer with no working scores M1A1)
(c)
For 1
B1
(d)
0.6
M1
Allow equivalents eg
a
+
b
=
Allow equivalents eg 5
+
6
=
3.35
M1
a
b
Correct method to eliminate a or b (implied by a correct value for a or b)
M1
This mark can still be scored even if the method leads to a value of a or b which is not a probability.
May see a = c – 0.4 to eliminate a or b = 1 – c used to eliminate b
A complete method for finding the value of c (condone using any value of a and b for this mark)
M1
A1
0.65 oe
(e)
M1
For the product of 3 probabilities
0.0675 or exact equivalent fraction eg 27
A1
400
(f)
Identifying that if Jessie scores 2, there is only one spin or the 4 correct possibilities only
M1
At least 3 correct nonzero probability products ft their a and b (an answer of 0.195 scores M0M1A0)
M1
A1
0.17
104.9 100
−
(
)
P
>
104.9
=
P
>
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