Mathematics · May / June 2023 · Paper 61 · Question 2

YMA01

As printed in the paper

The question as text

2. Two students, Olive and Shan, collect data on the weight, w grams, and the tail length, t cm, of 15 mice. Olive summarised the data as follows Stt = 5.3173 ∑w 2 = 6089.12 ∑tw = 2304.53 ∑w = 297.8 ∑t = 114.8

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Show the answer and mark scheme

2(a)[3]
297.8 114.8
×
297.8
2
S
=
2304.53
−
or
S
=
6089.12
−
tw
15
ww
15
awrt 25.4
S
tw =
25.367...
awrt 177
S
ww =
176.797
"25.367"
2(b)[2]
r =
5.3173 "176.797..."
×
= 0.82735…. awrt 0.827 or 0.828
2(c)[3]
"25.367..."
[
]
b =
=
4.77065...
5.3173
297.8
"25.367" 114.8
[
]
a =
−
×
= −
16.658....
15
5.3173
15
b = 4.771 or better or a = –16.66 or better seen and
= −
16.7 4.77 *
+
w
t
(d) [On average,] for each cm/1 cm of tail length/t the weight/w increases by 4.77 g/grams
16.7
2(d)[1]
297.8 114.8
×
297.8
2
S
=
2304.53
−
or
S
=
6089.12
−
tw
15
ww
15
awrt 25.4
S
tw =
25.367...
awrt 177
S
ww =
176.797
"25.367"
r =
5.3173 "176.797..."
×
= 0.82735…. awrt 0.827 or 0.828
"25.367..."
[
]
b =
=
4.77065...
5.3173
297.8
"25.367" 114.8
[
]
a =
−
×
= −
16.658....
15
5.3173
15
b = 4.771 or better or a = –16.66 or better seen and
= −
16.7 4.77 *
+
w
t
(d) [On average,] for each cm/1 cm of tail length/t the weight/w increases by 4.77 g/grams
16.7
[
]
4.77 2
[
9.54
]
or [
]
[
]
w = −
16.7
+
4.77
×
2
= −
7.16
or
×
=
t =
=
3.5
or sd = awrt 0.6
4.77
[
]
7.16
w = −
or 9.54 < 16.7 or 2 < 3.5 which is negative/weight cannot be negative
or for sd extrapolation since a 2 cm tail is (approx 9 sd)/(more than 3 sd) from the mean
0.827
(
)
2
+
10
= −
16.7
+
4.77
+
6
oe
y
x
Notes
(a)
M1 for a correct expression for Stw or Sww
A1 awrt 25.4
A1 awrt 177
(b)
M1 for a valid attempt at r with their Stwnot equal to 2304.53 and Sww not equal to 6089.12
A1 (M2 on epen) awrt 0.827 or awrt 0.828
1st M1 for a correct method to find the value of b
(c)
2nd M1 ft their b. For a correct method to find a. Minimum shown
[
]
=
awrt 19.9
−
"their " awrt 7.65
×
= −
16.658
a
b
A1* Both method marks must be awarded, equation stated (no fractions) and sight of (4.771 or better)
or (−16.66 or better)
(d)
B1 For a suitable contextual comment that implies that as length increases by 1 cm weight increases by
4.77g. Allow multiples eg each 10 cm increase in tail length weight increases by 47.7g Allow in terms
of t and w
M1 for a correct method to calculate the value of w (condone if written as a fraction) or
(e)
[
]
4.77 2
×
=
9.54
or correct method to find tail length when w = 0 or sd = awrt 0.6
A1 Method mark must be awarded. For –7.16 or 9.54 < 16.7 or 2 < 3.5 with a relevant explanation
stating that weight is negative. If sd = awrt 0.6 is given allow extrapolation since a 2 cm tail is (approx
9 sd)/(more than 3 sd) from the mean.
B1ft follow through their answer to (b)
(f)
(g)
−
16.7
+
4.77(
x
+
6)
B1 ISW no need to be simplified. Allow equivalent eg
=
−
5
The correct
y
2
simplified equation is
=
2.385
+
0.96
allow awrt 2.39 and 0.96 – 0.98
y
x
3711
[
]
2(e)[2]
[
]
4.77 2
[
9.54
]
or [
]
[
]
w = −
16.7
+
4.77
×
2
= −
7.16
or
×
=
t =
=
3.5
or sd = awrt 0.6
4.77
[
]
7.16
w = −
or 9.54 < 16.7 or 2 < 3.5 which is negative/weight cannot be negative
or for sd extrapolation since a 2 cm tail is (approx 9 sd)/(more than 3 sd) from the mean
2(f)[1]
0.827
(
)
2(g)[1]
2
+
10
= −
16.7
+
4.77
+
6
oe
y
x
Notes
(a)
M1 for a correct expression for Stw or Sww
A1 awrt 25.4
A1 awrt 177
(b)
M1 for a valid attempt at r with their Stwnot equal to 2304.53 and Sww not equal to 6089.12
A1 (M2 on epen) awrt 0.827 or awrt 0.828
1st M1 for a correct method to find the value of b
(c)
2nd M1 ft their b. For a correct method to find a. Minimum shown
[
]
=
awrt 19.9
−
"their " awrt 7.65
×
= −
16.658
a
b
A1* Both method marks must be awarded, equation stated (no fractions) and sight of (4.771 or better)
or (−16.66 or better)
(d)
B1 For a suitable contextual comment that implies that as length increases by 1 cm weight increases by
4.77g. Allow multiples eg each 10 cm increase in tail length weight increases by 47.7g Allow in terms
of t and w
M1 for a correct method to calculate the value of w (condone if written as a fraction) or
(e)
[
]
4.77 2
×
=
9.54
or correct method to find tail length when w = 0 or sd = awrt 0.6
A1 Method mark must be awarded. For –7.16 or 9.54 < 16.7 or 2 < 3.5 with a relevant explanation
stating that weight is negative. If sd = awrt 0.6 is given allow extrapolation since a 2 cm tail is (approx
9 sd)/(more than 3 sd) from the mean.
B1ft follow through their answer to (b)
(f)
(g)
−
16.7
+
4.77(
x
+
6)
B1 ISW no need to be simplified. Allow equivalent eg
=
−
5
The correct
y
2
simplified equation is
=
2.385
+
0.96
allow awrt 2.39 and 0.96 – 0.98
y
x
3711
[
]

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