Mathematics · Oct / Nov 2025 · Paper 61 · Question 2
YMA01
As printed in the paper
The question as text
2. The weights, to the nearest kilogram, of 35 rugby players from club A are summarised in the stem and leaf diagram below. Weight (kg) Totals Key: 7 | 2 represents 72 kg 7 2 (1) 8 4 9 (2) 9 0 0 1 1 5 5 5 6 7 (9) 10 0 0 4 4 5 5 7 8 9 9 (10) 11 1 1 2 2 2 2 5 (7) 12 0 2 4 6 7 (5) 13 9 (1)
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Show the answer and mark scheme
2(a)[2]
[LQ =] 95
[UQ =] 112
2(b)[3]
−
×
−
or "112" 1.5 ("112" "95")
+
×
−
"95" 1.5 ("112" "95")
−
×
−
and "112" 1.5 ("112" "95")
+
×
−
"95" 1.5 ("112" "95")
So only outlier is 139
Club A
2(c)[4]
60
70
80
90
100
140
110
120
130
2(d)[2]
(Club B has) positive (skew)
So club B weights are more varied above the median (than below the
median)
Notes
(a)
B1
correct lower quartile 95 cao
B1
correct upper quartile 112 cao
a correct method to find either the lower or upper outlier boundary ft their
(b)
M1
values from (a)
correct attempt to find both the lower and upper outlier boundary ft their
values from (a)
A1ft
this mark is for both correct ft expressions which need not be evaluated but
may be implied by both of 69.5 and 137.5
A1
(dep on M1 only) for identifying 139 as an outlier
(c)
M1
for a box with 1 upper whisker and 1 lower whisker
for lower whisker plotted at 72 and upper whisker plotted at 127 or ‘137.5
M1
(condone 137 to 138)’
M1
for quartiles plotted at ‘Q1’, 105 and ‘Q3’
for a fully correct boxplot with 1 outlier marked at 139 only ft ‘Q1’, ‘Q3’ and
A1ft
‘137.5’
for identifying positive (skew) ignore comments about the skewness of
(d)
M1
Club A
positive correlation is M0
for a correct contextual interpretation of positive skew indicating that there
A1
is greater variation above the median
1
(
)
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