Mathematics · Oct / Nov 2024 · Paper 61 · Question 5
YMA01
As printed in the paper
The question as text
5. Frequency density 0 10 20 30 40 50 Time in hours The histogram shows the number of hours worked in a given week by a group of 64 freelance photographers.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[1]
[Time is] continuous
16 photographers = 160 small squares or 64 photographers = 640 small squares
or 16
or 64
or 160
or 640
5(b)[3]
160 =
0.1
640 =
0.1
16 =
10
64 =
10
Frequency density = 1.6 or Correct scale on the frequency density axis
x =
16
oe or (20 12)
−
(24
−
20)
×
16
+
×
14
240
160
10
5
= 24
5(c)[2]
Using n
Using n + 1
(32
21)
(32.5
21)
−
−
[
=
]20
+
×
(25
−
20)
oe
[
Q =
]
20
+
×
(25
−
20)
oe
Q
2
35
−
21
2
35
−
21
35
−
32
35
−
32.5
or
25
(
25
20
)
oe
or
25
(
25
20
)
oe
−
×
−
−
×
−
35
21
35
21
−
−
= awrt 23.9
= awrt 24.1
5(d)[2]
Mean = Median or Mean ≈ Median
e.g. Appropriate decision. Consistent with expectation for a normal distribution.
Notes
(a)
B1
Allow not discrete
For establishing a ratio between photographers and area or calculating frequency density (may be
(b)
M1
implied by 2nd M1)
M1
For a correct ratio or expression using areas for photographers from 12 to 24
A1
Cao
(c)
M1
For a correct method to find median using either n or n + 1
A1
awrt 23.9 or awrt 24.1 if using n + 1
For a correct comment about mean and median ft their median
(d)
M1
Allow mean is close to median to imply mean ≈ median
Ignore any comments made about the shape of the histogram
For a correct compatible comment about Charlie’s decision ft their median
If Mean = Median or Mean ≈ Median, then the decision should be that a normal distribution is
A1ft
suitable [due to symmetry]
If Mean < Median or Mean > Median or mean ≠ median, then the decision should be that a
normal distribution is not suitable [due to the skew in the data]
[
]
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