Mathematics · May / June 2024 · Paper 61 · Question 3

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As printed in the paper

The question as text

3. The lengths, x mm, of 50 pebbles are summarised in the table below. Length Frequency 20  x < 30 2 30  x < 32 16 32  x < 36 20 36  x < 40 8 40  x < 45 3 45  x < 50 1 A histogram is drawn to represent these data. The bar representing the class 32  x < 36 is 2.5 cm wide and 7.5 cm tall.

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3(a)[3]
Width = 1.25 [cm]
18.75 cm2 for freq of 20 so 18.75 16
20 ×
=
15
×
=
15
cm2 for a frequency of 16 or
w h
or fd = 5
[
h =
15 1.25
÷
or
h =
8
÷×
5 7.5
=] 12 (cm)
7
7.5
3(b)[2]
Q =
[32 ]
+
×
4
or using n + 1 gives
Q =
[32 ]
+
×
4
2
20
2
20
=33.4 (
n + gives 33.5)
1
104
[
]
(
)
[
]
[
]
2.08
w−
20
=
10 104
×
=
1040
w =
10 104
×
+
50 20
×
=
2040
3(c)[2]
y =
=
or
50
"1040"
or "2040"
w =
10 "2.08" 20
×
+
= 40.8*
+
20
=
40.8
=
40.8
50
50
233.54
861
[
]
(
)
Variance of
=
−
"2.08"
2
=
=
0.3444
or 10 × sd of y = sd of w
y
50
2500
3(d)[4]
(
)
(
)
or
100 233.54
×
=
w
2
−
40 "2040" 50 400
×
+
×
w
2
=
84954
oe
"84954"
861
[
]
Variance of
"0.3444" 100
40.8
2
34.44
w =
×
or
−
=
=
50
25
861
or
sd of
y
=
"0.3444"
=
=
0.5868...
50
861
sd of w = "0.3444" 100
×
or "34.44" or
10 "
×
"
50
= 5.868… awrt 5.87
(e)(i)
The mean would not change (as 40.8 is the mean)
(ii)
The standard deviation would decrease (as 40.8 is in the middle so data closer together)
Both correct with a correct reason for why the standard deviation decreases
Notes
(a)
for width = 1.25 no need for units
B1
for sight of 15 or "their
w
" "their "
×
h
=
15
or fd = 5 May be implied by h = 12
M1
for height = 12 no need for units
A1
for 7
20 ×or 13
m −
32
4
oe or 36
−
4
20 ×or
m
oe (allow 25.5 rather than 25)
(b)
M1
4
4
=
=
25 18
−
20
38
−
25
20
33.4 or if using (
)1
n +
33.5
A1
(
)
for a correct method to find the mean of y or
w −
20
or
w
(c)
M1
(10 104
×
+
k
where
k ≠
20 50
×
is M0)
for a correct method to find mean of w which leads to 40.8
A1*
for a correct method to find the Variance of y or writing/using 10 ×sd of y = sd of w or a correct
(d)
M1
2
equation to find
w
for a correct method to find the Variance of w or sd of y ft their Var(y)
M1
M1
for a correct method to find the sd of w ft their Var(w)
awrt 5.87 NB an exact answer
861
scores A0
A1
5
(e)(i)
for no change (No reason needed) Allow mean = 40.8 to imply no change
B1
(ii)
for sd decreases/be smaller/go down (condone Var decreases) (No reason needed)
B1
Both previous B1 awarded. For a correct reason for the sd decreasing Allow (
)
x
−
x
ddB1
(
)
2
Allow
x
−
x
doesn’t change and n increases. Allow the values would be more
concentrated about the mean
1456.8 713.2
713.2
2
×
3(e)[3]
Width = 1.25 [cm]
18.75 cm2 for freq of 20 so 18.75 16
20 ×
=
15
×
=
15
cm2 for a frequency of 16 or
w h
or fd = 5
[
h =
15 1.25
÷
or
h =
8
÷×
5 7.5
=] 12 (cm)
7
7.5
Q =
[32 ]
+
×
4
or using n + 1 gives
Q =
[32 ]
+
×
4
2
20
2
20
=33.4 (
n + gives 33.5)
1
104
[
]
(
)
[
]
[
]
2.08
w−
20
=
10 104
×
=
1040
w =
10 104
×
+
50 20
×
=
2040
y =
=
or
50
"1040"
or "2040"
w =
10 "2.08" 20
×
+
= 40.8*
+
20
=
40.8
=
40.8
50
50
233.54
861
[
]
(
)
Variance of
=
−
"2.08"
2
=
=
0.3444
or 10 × sd of y = sd of w
y
50
2500
(
)
(
)
or
100 233.54
×
=
w
2
−
40 "2040" 50 400
×
+
×
w
2
=
84954
oe
"84954"
861
[
]
Variance of
"0.3444" 100
40.8
2
34.44
w =
×
or
−
=
=
50
25
861
or
sd of
y
=
"0.3444"
=
=
0.5868...
50
861
sd of w = "0.3444" 100
×
or "34.44" or
10 "
×
"
50
= 5.868… awrt 5.87
(e)(i)
The mean would not change (as 40.8 is the mean)
(ii)
The standard deviation would decrease (as 40.8 is in the middle so data closer together)
Both correct with a correct reason for why the standard deviation decreases
Notes
(a)
for width = 1.25 no need for units
B1
for sight of 15 or "their
w
" "their "
×
h
=
15
or fd = 5 May be implied by h = 12
M1
for height = 12 no need for units
A1
for 7
20 ×or 13
m −
32
4
oe or 36
−
4
20 ×or
m
oe (allow 25.5 rather than 25)
(b)
M1
4
4
=
=
25 18
−
20
38
−
25
20
33.4 or if using (
)1
n +
33.5
A1
(
)
for a correct method to find the mean of y or
w −
20
or
w
(c)
M1
(10 104
×
+
k
where
k ≠
20 50
×
is M0)
for a correct method to find mean of w which leads to 40.8
A1*
for a correct method to find the Variance of y or writing/using 10 ×sd of y = sd of w or a correct
(d)
M1
2
equation to find
w
for a correct method to find the Variance of w or sd of y ft their Var(y)
M1
M1
for a correct method to find the sd of w ft their Var(w)
awrt 5.87 NB an exact answer
861
scores A0
A1
5
(e)(i)
for no change (No reason needed) Allow mean = 40.8 to imply no change
B1
(ii)
for sd decreases/be smaller/go down (condone Var decreases) (No reason needed)
B1
Both previous B1 awarded. For a correct reason for the sd decreasing Allow (
)
x
−
x
ddB1
(
)
2
Allow
x
−
x
doesn’t change and n increases. Allow the values would be more
concentrated about the mean
1456.8 713.2
713.2
2
×

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