Mathematics · Oct / Nov 2022 · Paper 61 · Question 5
YMA01
As printed in the paper
The question as text
5. The weights, W grams, of kiwi fruit grown on a farm are normally distributed with mean 80 grams and standard deviation 8 grams. The table shows the classifications of the kiwi fruit by their weight, where k is a positive constant. Small Large Tiny Petite Extra Jumbo Mega w < 66 66 w < 70 70 w < 80 80 w < k w k One kiwi fruit is selected at random from those grown on the farm.
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5(a)[3]
W
Z
8
=
P(
Z
>−
1.25) or P(
Z
<
1.25)
=
0.8944
awrt 0.894
5(b)[4]
P(
<
)
=
0.85 or P(
>
)
=
0.15
W
k
W
k
k −
80
±
=
1.0364
8
k = 88.29… awrt 88.3
66
−
80
P(
66)
P
[
1.75]
[= 0.0401 (calc 0.040059…)]
5(c)[5]
W
<
=
Z
<
= −
8
(
)
0.25 P(
×
Z
<−
1.75)
=
0.010025 calc0.0100147...
or 0.25 (1 P(
×
−
Z
<
1.75))
dM1
y −
80
= –2.32(63)
8
y = 61.389… awrt 61.4 A1
Notes
(a) M1 for standardising with 70, 80 and 8 (allow ±)
1st A1 z = ±1.25
2nd A1 awrt 0.894 (calc 0.894350…) NB do not ISW so an answer of 0.1056 is A0
(b) 1st B1 for either correct probability statement. Allow
„
for < and
…
for >
(may be implied by z = awrt 1.04)
M1 standardising with 80, 8 and equating to z, where 1 < |z| < 2
2nd B1 z = ±1.0364 or better (calc 1.036432…)
A1 awrt 88.3 (calc 88.291459…)
NB awrt 88.3 implies 1st B1 and M1 but not the 2nd B1 they could get B1M1B0A1
(Answer only 88.291 to 88.292 scored 4 out of 4)
(c) 1st M1 standardising with 66, 80 and 8 (allow ±) or seeing awrt 0.0401 in working
2nd dM1 (dep on 1st M1) 0.25 "their P(
×
<−
1.75)"
or 0.0401 – 0.030075 or seeing
Z
[
]
0.75 0.0401 0.9599
×
+
= 0.9899…
3rd M1 for standardising and equating to z, where |z| > 2
1st A1 correct standardisation equation with compatible signs and 2.32 „ |z| „ 2.34
2nd A1 awrt 61.4 (allow awrt 61.3)
6. (a)(i) [P(A) =] 0.25
(ii) [
]
P(
A B = 1
|
)
(iii) [
]
P(
A C = 0
|
)
q
3
r =
(b)
+
5
q
7
0.13
+
+
=
p
s
10
p
+
q
+ + +
r
s
0.12
+
0.13
=
1
Solving simultaneously to get
3
q
0.3
0.12
=
or 0.3
=
0.12 1.5r
+
+
r
or
=
+
q
+
2
q
oe
0.3
−
0.12
5
3
q = 0.108
r = 0.072
5
(c)
=
0.13
+
0.12
+
'0.072'
+ oe
s
8
s = 0.303 A1
Notes
(a)(i) B1 0.25 oe
(ii)
B1 1 cao
(iii)
B1 0 cao
C D = 3
(b) 1st M1 correct expression for P(
|
)
5 . Allow P(D) for q + r
2nd M1 correct expression for P(
= 7
′
∩
′
)
10
B
D
NB 3
3rd M1 A correct equation or use of sum of probabilities = 1
=
0.12
+
q
+ is
must imply correct equation eg may ft their
10
= 7
2ndM1 3rdM1
P(
B
′
∩
D
′
)
10 Implied by
q
+
r
=
0.18
or P(D) = 0.18
4th dM1 (dep on all 3 previous M1 being awarded) solving to obtain a correct equation in a
single variable. Implied by a correct value for q or r
1st A1 q = 0.108 or 27
250 oe
2nd A1 r = 0.072 or 9
125 oe
5
C′
(c) M1 correct expression for
P(
B
∪
)
=
ft their value for r. Allow use of the letter r
8
eg 5
=
0.13
+
0.12
+ + oe We will condone values of r outside the range 0 < r <1
r
s
8
A1 s = 0.303 oe
0.1
=
1
8
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