Mathematics · May / June 2023 · Paper 61 · Question 7
YMA01
As printed in the paper
The question as text
7. A machine squeezes apples to extract their juice. The volume of juice, J ml, extracted from 1 kg of apples is modelled by a normal distribution with mean μ and standard deviation σ Given that μ = 500 and σ = 25 use standardisation to
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7(a)(i)[2]
510
−
500
(
)
P
J >
510
=
P
>
or P(Z > 0.4)
Z
25
= 1 – 0.6554
0.3446
*
7(a)(ii)[3]
d −
500
1.4
= −
(calc –1.3997…)
25
d = 465 (calc 465.007)
7(b)[2]
(
1 0.3446
)
5
−
= 0.1209… awrt 0.121
7(c)[7]
r
−
520
= −
1.0364
k
3
−
800
−
520
r
=
2.5758
k
r
−
520
3
r
−
1320
(
)
−
240
=
3
×−
"1.0364 "
−
"2.5758"
or
=
oe
k
k
" 1.0364"
−
"2.5758"
42.216...
k =
awrt 42
r =
476.246...
awrt 476
Notes
(a)(i)
M1 for standardising using 500 and 25. Allow for 0.4
A1* M1 must be awarded. For 1 – 0.6554 = 0.3446 or using calc 0.34457… = 0.3446 or better
(ii)
M1 correct standardisation using 500 and 25 equated to a z value where1
<
<
2
z
B1 correct expression with compatible signs eg 500
−
d
=
1.4
(calc 1.3997…) or allow incompatible
25
signs with 500
−
("535" 500)
−
SC 510
−
d
=
1.4
(calc 1.3997…) can get M0B1A0
25
dA1 dependent on M1 awarded for 465 or 465.007...
(b)
M1 for ( )
5
where 0 < p < 1
p
A1 awrt 0.121
(c)
r
−
520
M1
= z value where |z| > 1
k
−
520
r
(calc 1.036433...) (signs must be compatible)
1st A1
=
awrt
−
1.0364
k
2nd M1 3
−
800
−
520
r
= z value where |z| > 2
k
2nd A1 3
−
800
−
520
r
awrt 2.5758
(calc 2.5758293…) (signs must be compatible)
=
k
3rd M1 (dep on both Ms) for forming a correct equation in k or r only using their z values. ISW once
correct equation seen eg –5.685k = –240 or (
3
1.0364
520
)
800
520
−
k
+
−
−
=
2.5758
Implied by r
k
and k correct
3rdA1 for awrt 42
4th A1 for awrt 476 Must come from equations with compatible signs
NB awrt 476 and awrt 42 does not mean full marks. They could get M1A0M1A0 M1A1A1 if they do
not have accurate z values
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