Mathematics · Oct / Nov 2023 · Paper 61 · Question 5
YMA01
As printed in the paper
The question as text
5. The weights, X grams, of a particular variety of fruit are normally distributed with 2) X ~ N(210, 25 A fruit of this variety is selected at random.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[2]
N(210,25 )
2
X
240
−
210
[
]
P(
X
<
240)
=
P
Z
<
=
P(
Z
<
1.2)
25
= 0.8849*
190
−
210
[
]
P(190
240)
0.8849
P
0.8849
P(
0.8)
5(b)[2]
<
X
<
=
−
Z
<
=
−
Z
<−
25
0.8849
−
0.2119
=
0.673
awrt
0.673
210
+
−
210
or 210
−
−
210
k
k
5(c)[3]
=
1.96
= −
1.96
25
25
k =
49
awrt 49 A1
S
−
210
(
)
5(d)[3]
P
<
=
0.15
= −
1.0364
X
S
25
S =
184.09
awrt 184 A1
5(e)[5]
N( ,
2
)
Y
μ
152
−
μ
P
(
<
152
)
=
0.05
= −
1.6449
Y
180
−
μ
(
)
P
Y
>
180
=
0.40
=
0.2533
28
=
1.8982
=
14.75...
and
μ=
176.26...
Notes
(a)
M1
For standardising using 240, 210 and 25
Cao As the answer is given then no incorrect working should be seen
A1*
For standardising using 190/230, 210 and 25 and subtracting from 0.8849 May be implied by
(b)
M1
(1.2)
(0.8) 1
−or 0.8849 + 0.7881 – 1
+
awrt 0.673
A1
(c)
For standardising and setting equal to a z value, where 1.9
<
z
<
2
M1
For
z =
1.96
or better
B1
A1
awrt 49
For standardising using S (allow any letter) and setting equal to a z value, where 1
<
(d)
M1
B1
For z = 1.0364
awrt 184
A1
For a correct method to form an equation in μ and set equal to a z value, where
(e)
M1
−
1.6
<
<−
1.7
or 0.2
<
<
0.3
(Signs must be compatible)
z
z
For a correct equation for (
)
A1
P
Y <
152
For a correct equation for (
)
P
Y >
180
A1
Dependent on previous M mark. For solving the 2 equations simultaneously. If answers are
dM1
incorrect then working must be shown. May be implied by
=
awrt 14.8
and
μ=
For
=
awrt 14.8
and
μ=
awrt 176
A1
(
)
1.4(1.2
0.2 ) 1.5
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