28 The table shows information about the weights of 120 oranges.
Weight (\(w\) grams)
Frequency
\(50 \lt w \leqslant 100\)
34
\(100 \lt w \leqslant 150\)
29
\(150 \lt w \leqslant 200\)
27
\(200 \lt w \leqslant 250\)
19
\(250 \lt w \leqslant 300\)
11
(a) Find the class interval that contains the median. (1)
(b) Calculate an estimate for the mean weight of the 120 oranges. Give your answer correct to 3 significant figures. (3)
Mark scheme (a)
Answer
Mark
Mark scheme
\(100 \lt w \leqslant 150\)
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
152
M1
for finding 5 products within the interval (including end points) with not more than one error, may be seen near table, eg \(75 \times 34\ (= 2550)\), \(125 \times 29\ (= 3625)\), \(175 \times 27\ (= 4725)\), \(225 \times 19\ (= 4275)\), \(275 \times 11\ (= 3025)\) or for 18200
M1
for \(\Sigma fx \div \Sigma f\) eg \((\text{``}2550\text{''} + \text{``}3625\text{''} + \text{``}4725\text{''} + \text{``}4275\text{''} + \text{``}3025\text{''}) \div 120\) or \(\text{``}18200\text{''} \div 120\)
A1
for answer in the range 151 to 152
Additional guidance
do not award this mark if the final answer comes from an alternative incorrect method, eg \(120 \div 5\ (= 24)\) or \(\Sigma mp \div \Sigma f\ (875 \div 120\ (= 7.29\ldots))\) or \(\Sigma mp \div 5\ (875 \div 5\ (= 175))\)
Min \(fx\)
Max \(fx\)
1700
3400
2900
4350
4050
5400
3800
4750
2750
3300
15200
21200
\(\Sigma fx\) must come from 5 products, \(fx\) within intervals (including end points)
Correct midpoints must be used for the award of the A mark
22 Seija works at a weather station. The table gives information about the temperature, \(T\) °C, at midday for each of 50 cities in the UK on Tuesday.
Temperature (\(T\) °C)
Frequency
\(10 \lt T \leqslant 15\)
2
\(15 \lt T \leqslant 20\)
8
\(20 \lt T \leqslant 25\)
13
\(25 \lt T \leqslant 30\)
21
\(30 \lt T \leqslant 35\)
6
(a) Calculate an estimate for the mean temperature. (3)
Seija says,
“The median temperature is 22.5 °C because 22.5 is the middle number in the middle group.”
(b) Is Seija correct? Give a reason for your answer. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
24.6
M1
for finding 5 products within intervals (including end points) with not more than one error, may be seen near table. eg \(2 \times 12.5\ (= 25)\), \(8 \times 17.5\ (= 140)\), \(13 \times 22.5\ (= 292.5)\), \(21 \times 27.5\ (= 577.5)\), \(6 \times 32.5\ (= 195)\) or for 1230
M1
for \(\Sigma fx \div \Sigma f\) eg \((\text{``}25\text{''} + \text{``}140\text{''} + \text{``}292.5\text{''} + \text{``}577.5\text{''} + \text{``}195\text{''}) \div \text{``}50\text{''}\) or \(\text{``}1230\text{''} \div \text{``}50\text{''}\)
A1
for 24.6 oe
Additional guidance
Min \(fx\)
Max \(fx\)
20
30
120
160
260
325
525
630
180
210
\(\Sigma fx\) must come from 5 products, \(fx\) within intervals (including end points)
Mark scheme (b)
Answer
Mark
Mark scheme
No, with reason
C1
for No and reason
Acceptable No, the median is in the interval \(25 \lt T \leqslant 30\) No, the median is in the group containing the 25(.5)th temperature No, she did not take into account frequency No, the frequencies are not the same for each group.
Not acceptable No, the median is 27.5 No, the median is higher than 22.5 \(25 \lt T \leqslant 30\) Yes, …
Additional guidance
Any incorrect statement as part of a correct response can be ignored unless it contradicts the statement.
3 Seija works at a weather station. The table gives information about the temperature, \(T\) °C, at midday for each of 50 cities in the UK on Tuesday.
Temperature (\(T\) °C)
Frequency
\(10 \lt T \leqslant 15\)
2
\(15 \lt T \leqslant 20\)
8
\(20 \lt T \leqslant 25\)
13
\(25 \lt T \leqslant 30\)
21
\(30 \lt T \leqslant 35\)
6
(a) Calculate an estimate for the mean temperature. (3)
Seija says,
“The median temperature is 22.5 °C because 22.5 is the middle number in the middle group.”
(b) Is Seija correct? Give a reason for your answer. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
24.6
M1
for finding 5 products within intervals (including end points) with not more than one error, may be seen near table. eg \(2 \times 12.5\ (= 25)\), \(8 \times 17.5\ (= 140)\), \(13 \times 22.5\ (= 292.5)\), \(21 \times 27.5\ (= 577.5)\), \(6 \times 32.5\ (= 195)\) or for 1230
M1
for \(\Sigma fx \div \Sigma f\) eg \((\text{``}25\text{''} + \text{``}140\text{''} + \text{``}292.5\text{''} + \text{``}577.5\text{''} + \text{``}195\text{''}) \div \text{``}50\text{''}\) or \(\text{``}1230\text{''} \div \text{``}50\text{''}\)
A1
for 24.6 oe
Additional guidance
Min \(fx\)
Max \(fx\)
20
30
120
160
260
325
525
630
180
210
\(\Sigma fx\) must come from 5 products, \(fx\) within intervals (including end points)
Mark scheme (b)
Answer
Mark
Mark scheme
No, with reason
C1
for No and reason
Acceptable examples No, the median is in the interval \(25 \lt T \leqslant 30\) No, the median is in the group containing the 25(.5)th temperature No, she did not take into account frequency No, the frequencies are not the same for each group.
Not acceptable examples No, the median is 27.5 No, the median is higher than 22.5 \(25 \lt T \leqslant 30\) Yes, …
Additional guidance
Any incorrect statement as part of a correct response can be ignored unless it contradicts the statement.
for fully correct polygon with points plotted at the midpoints
(B1
for points plotted correctly but not joined by straight lines or joining points at correct heights consistently within intervals including plotting at end values or correct frequency polygon with one point incorrect or correct frequency polygon with first and last points joined directly)
Additional guidance
For B2: Joining must be with line segments
For B1: for example, at 10, 20, 30,...or at 20, 30, 40,... Ignore any histogram drawn and any part of frequency polygon outside range of first and last points plotted
3 The table shows information about the heights of 80 plants.
Height (\(h\) cm)
Frequency
\(10 \lt h \leqslant 20\)
7
\(20 \lt h \leqslant 30\)
13
\(30 \lt h \leqslant 40\)
14
\(40 \lt h \leqslant 50\)
12
\(50 \lt h \leqslant 60\)
16
\(60 \lt h \leqslant 70\)
18
(a) Find the class interval that contains the median. (1)
(b) On the grid, draw a frequency polygon for the information in the table.(2)
Mark scheme (a)
Answer
Mark
Mark scheme
\(40 \lt h \leqslant 50\)
B1
accept 40 – 50 oe
Mark scheme (b)
Answer
Mark
Mark scheme
polygon drawn
B2
for fully correct polygon with points plotted at the midpoints
(B1
for points plotted correctly but not joined by straight lines or joining points at correct heights consistently within intervals including plotting at end values or correct frequency polygon with one point incorrect or correct frequency polygon with first and last points joined directly)
Additional guidance
Joining must be with line segments (15, 7), (25, 13) (35, 14), (45, 12) (55, 16), (65, 18)
For example, at 10, 20, 30,… or at 20, 30, 40,…
Ignore any histogram drawn and any part of frequency polygon outside range of first and last points plotted
27 The table shows information about the weekly earnings of 20 people who work in a shop.
Weekly earnings (£\(x\))
Frequency
\(150 \lt x \leqslant 250\)
1
\(250 \lt x \leqslant 350\)
11
\(350 \lt x \leqslant 450\)
5
\(450 \lt x \leqslant 550\)
0
\(550 \lt x \leqslant 650\)
3
(a) Work out an estimate for the mean of the weekly earnings. (3)
Nadiya says,
“The mean may not be the best average to use to represent this information.”
(b) Do you agree with Nadiya? You must justify your answer. (1)
Mark scheme (a)
Answer
Mark
Notes
365
M1
\(fx\) with \(x\) consistent within intervals eg \(200 \times 1\), \(300 \times 11\), \(400 \times 5\), \(500 \times 0\), \(600 \times 3\), if 200, 3300, 2000, 0, 1800 are seen without working then condone 1 error
M1
(dep) \(\Sigma fx \div \Sigma f\) eg \(\text{``}7300\text{''} \div 20\)
19 The table shows information about the heights of 80 children.
Height (\(h\) cm)
Frequency
\(130 \lt h \leqslant 140\)
4
\(140 \lt h \leqslant 150\)
11
\(150 \lt h \leqslant 160\)
24
\(160 \lt h \leqslant 170\)
22
\(170 \lt h \leqslant 180\)
19
(a) Find the class interval that contains the median. (1)
(b) Draw a frequency polygon for the information in the table.(2)
Mark scheme (a)
Answer
Mark
Notes
\(160 \lt h \leqslant 170\)
B1
correct class interval
Mark scheme (b)
Answer
Mark
Notes
Line segments joining the points (135, 4), (145, 11), (155, 24), (165, 22) and (175, 19)
C2
for fully correct frequency polygon
[C1
for points plotted correctly at midpoints of intervals OR joining points with line segments at the correct heights and consistent within the intervals (including end values) OR correct frequency polygon with one point incorrect OR correct frequency polygon with first and last point joined] NB: ignore any histogram drawn and any part of frequency polygon outside range of first and last points plotted
5 The table shows information about the weekly earnings of 20 people who work in a shop.
Weekly earnings (£\(x\))
Frequency
\(150 \lt x \leqslant 250\)
1
\(250 \lt x \leqslant 350\)
11
\(350 \lt x \leqslant 450\)
5
\(450 \lt x \leqslant 550\)
0
\(550 \lt x \leqslant 650\)
3
(a) Work out an estimate for the mean of the weekly earnings. (3)
Nadiya says,
“The mean may not be the best average to use to represent this information.”
(b) Do you agree with Nadiya? You must justify your answer. (1)
Mark scheme (a)
Answer
Mark
Notes
365
M1
\(fx\) with \(x\) consistent within intervals eg \(200 \times 1\), \(300 \times 11\), \(400 \times 5\), \(500 \times 0\), \(600 \times 3\), if 200, 3300, 2000, 0, 1800 are seen without working then condone 1 error
M1
(dep) \(\Sigma fx \div \Sigma f\) eg \(\text{``}7300\text{''} \div 20\)
1 The table shows information about the heights of 80 children.
Height (\(h\) cm)
Frequency
\(130 \lt h \leqslant 140\)
4
\(140 \lt h \leqslant 150\)
11
\(150 \lt h \leqslant 160\)
24
\(160 \lt h \leqslant 170\)
22
\(170 \lt h \leqslant 180\)
19
(a) Find the class interval that contains the median. (1)
(b) Draw a frequency polygon for the information in the table.(2)
Mark scheme (a)
Answer
Mark
Notes
\(160 \lt h \leqslant 170\)
B1
correct class interval
Mark scheme (b)
Answer
Mark
Notes
Line segments joining the points (135, 4), (145, 11), (155, 24), (165, 22) and (175, 19)
C2
for fully correct frequency polygon
[C1
for points plotted correctly at midpoints of intervals OR joining points with line segments at the correct heights and consistent within the intervals (including end values) OR correct frequency polygon with one point incorrect OR correct frequency polygon with first and last point joined] NB: ignore any histogram drawn and any part of frequency polygon outside range of first and last points plotted
20 The table gives information about the money, £\(A\), some people spent on an internet site one day.
Money spent (£\(A\))
Frequency
\(0 \lt A \leqslant 20\)
10
\(20 \lt A \leqslant 40\)
15
\(40 \lt A \leqslant 60\)
25
\(60 \lt A \leqslant 80\)
40
\(80 \lt A \leqslant 100\)
6
(a) On the grid, draw a frequency polygon for this information.(2)
(b) Write down the modal class interval. (1)
Mark scheme (a)
Answer
Mark
Notes
Frequency polygon
2
B2 correct frequency polygon (B1 for points plotted correctly but not joined or for points plotted at the correct heights, consistently placed within the class intervals (including ends) and joined or for an otherwise correct frequency polygon with no more than one point incorrect or correct frequency polygon with first and last points joined directly)
NB: ignore parts of graph drawn to the left of the 1st point or the right of the last point; ignore any histograms drawn.
M1 for \(fx\) consistently within interval including ends (allow 1 error) M1 consistently using appropriate midpoints M1 (dep on first M1) for \(\Sigma fx \div \Sigma f\) eg \(52.15 \div 35\) A1 cao
3 The table gives information about the money, £\(A\), some people spent on an internet site one day.
Money spent (£\(A\))
Frequency
\(0 \lt A \leqslant 20\)
10
\(20 \lt A \leqslant 40\)
15
\(40 \lt A \leqslant 60\)
25
\(60 \lt A \leqslant 80\)
40
\(80 \lt A \leqslant 100\)
6
(a) On the grid, draw a frequency polygon for this information.(2)
(b) Write down the modal class interval. (1)
Mark scheme (a)
Answer
Mark
Notes
Frequency polygon
2
B2 correct frequency polygon (B1 for points plotted correctly but not joined OR for points plotted at the correct heights, consistently placed within the class intervals (including ends) and joined OR for an otherwise correct frequency polygon with one point incorrect OR correct frequency polygon with first and last points joined directly)
NB: ignore parts of graph drawn to the left of the 1st point or the right of the last point; ignore any histograms drawn.
22 The table shows some information about the ages of 60 teachers.
Age (\(a\) years)
Frequency
\(20 \lt a \leqslant 30\)
6
\(30 \lt a \leqslant 40\)
16
\(40 \lt a \leqslant 50\)
14
\(50 \lt a \leqslant 60\)
22
\(60 \lt a \leqslant 70\)
2
(a) Write down the modal class interval. (1)
(b) Draw a frequency polygon for the information in the table.(2)
Mark scheme (a)
Answer
Mark
Notes
\(50 \lt a \leqslant 60\)
1
B1 for correctly identifying the modal class interval e.g. 50 – 60 oe
Mark scheme (b)
Answer
Mark
Notes
Polygon
2
B2 for fully correct frequency polygon - points plotted at the midpoint (B1 for all points plotted accurately but not joined with straight line segments or all points plotted accurately and joined with last joined to first to make a polygon or all points at the correct heights and consistently within or at the ends of the intervals and joined (can include joining last to first to make a polygon)
NB: ignore parts of graph drawn to the left of the 1st point or the right of the last point; ignore any histograms drawn.
6 The table shows some information about the ages of 60 teachers.
Age (\(a\) years)
Frequency
\(20 \lt a \leqslant 30\)
6
\(30 \lt a \leqslant 40\)
16
\(40 \lt a \leqslant 50\)
14
\(50 \lt a \leqslant 60\)
22
\(60 \lt a \leqslant 70\)
2
(a) Write down the modal class interval. (1)
(b) Draw a frequency polygon for the information in the table.(2)
Mark scheme (a)
Answer
Mark
Notes
\(50 \lt a \leqslant 60\)
1
B1 for correctly identifying the modal class interval e.g. 50 – 60 oe
Mark scheme (b)
Answer
Mark
Notes
Polygon
2
B2 for fully correct frequency polygon - points plotted at the midpoint (B1 for all points plotted accurately but not joined with straight line segments or all points plotted accurately and joined with last joined to first to make a polygon or all points at the correct heights and consistently within or at the ends of the intervals and joined (can include joining last to first to make a polygon)) NB: ignore parts of graph drawn to the left of the 1st point or the right of the last point; ignore any histograms drawn.
24 The table shows information about the heights of 50 trees.
Height (\(h\) metres)
Frequency
\(0 \lt h \leqslant 4\)
8
\(4 \lt h \leqslant 8\)
21
\(8 \lt h \leqslant 12\)
12
\(12 \lt h \leqslant 16\)
7
\(16 \lt h \leqslant 20\)
2
Draw a frequency polygon for the information in the table. (2)
Mark scheme
Answer
Mark
Notes
Polygon drawn
2
B2 for correct frequency polygon (B1 for points plotted at correct midpoints of intervals or joining points at correct heights consistently within intervals including plotting at end values or correct frequency polygon with one point incorrect or correct frequency polygon with first and last points joined directly)
NB ignore any histogram drawn and any part of frequency polygon outside range of first and last points plotted
M1 for \(fx\) with \(x\) consistently within interval including ends (allow 1 error) M1 for consistently using appropriate midpoints in \(fx\) M1 (dep on first M1) for \(\text{``}\Sigma fx\text{''} \div 50\) (or divided by \(\text{``}\Sigma f\text{''}\) if clearly calculated), eg \(396 \div 50\) A1 for 7.92 cao
M1 for finding at least 4 products ft consistently within interval (including end points) M1 (dep) for use of at least 4 correct midpoints. M1 (dep on 1st M) for \(\text{'}\Sigma ft\text{'} \div 40\) A1 for 116.25
M1 for \(fx\) consistently within interval including ends (allow 1 error) M1 consistently using appropriate midpoints M1 (dep on first M1) for \(\Sigma fx \div \Sigma f\) A1 for 18.2 cao
Mark scheme (b)
Working
Answer
Mark
Notes
\(0 \leqslant t \lt 10\) fd 0.8 \(10 \leqslant t \lt 15\) fd 3 \(15 \leqslant t \lt 20\) fd 2.2 \(20 \leqslant t \lt 30\) fd 1 \(30 \leqslant t \lt 50\) fd 0.3
Correct histogram
3
B3 fully correct histogram with vertical axis correctly scaled. (B2 for 4 correct blocks or 5 correct blocks with incorrect or no scale) (B1 for 2 correct blocks of different widths or any 3 correct blocks or correct FD values for at least 3 frequencies) eg fd of 0.8, 3, 2.2, 1, 0.3
M1 for finding \(fx\) with \(x\) consistent within intervals (including the end points) allow 1 error; implied by 820, 1020 M1 (dep) for use of all correct mid-interval values eg 920 M1 (dep on 1st M1) for \(\sum fx \div \sum f\) A1 for 18.4 oe
Mark scheme (c)
Answer
Mark
Notes
correct frequency polygon
2
B2 for fully correct frequency polygon - points plotted at the midpoint (B1 for all points plotted accurately but not joined with straight line segments) or all points plotted accurately and joined with last joined to first to make a polygon or all points at the correct heights and consistently within or at the ends of the intervals and joined (can include joining last to first to make a polygon) NB: ignore parts of graph drawn to the left of the 1st point or the right of the last point
8 Here are the test marks for the students in Mrs Potter’s maths class.
28
36
25
43
18
39
30
36
27
31
33
14
22
36
23
12
38
36
40
45
27
(a) Complete the frequency table for these marks.
Mark
Tally
Frequency
10 – 19
20 – 29
30 – 39
40 – 49
(3)
(b) Work out the number of these students getting less than 20 marks. (1)
Mark scheme (a)
Answer
Mark
Notes
3, 6, 9, 3
3
B3 for a table showing all 4 correct frequencies in the correct place. (condone the absence of or any incorrect tallies) [(B2 for 2 or 3 correct tallies or 2 or 3 correct frequencies even in the wrong columns) (B1 for 1 correct tally or 1 correct frequency even in the wrong column)] [SC: B2 for \(\frac{3}{21}\) and \(\frac{6}{21}\) and \(\frac{9}{21}\) and \(\frac{3}{21}\) shown in the frequency column]
12 The frequency table gives information about the times it took some office workers to get to the office one day.
Time (\(t\) minutes)
Frequency
\(0 \lt t \leqslant 10\)
4
\(10 \lt t \leqslant 20\)
8
\(20 \lt t \leqslant 30\)
14
\(30 \lt t \leqslant 40\)
16
\(40 \lt t \leqslant 50\)
6
\(50 \lt t \leqslant 60\)
2
(a) Draw a frequency polygon for this information.(2)
(b) Write down the modal class interval. (1)
One of the office workers is chosen at random.
(c) Work out the probability that this office worker took more than 40 minutes to get to the office. (2)
Mark scheme (a)
Working
Answer
Mark
Notes
Correct Frequency Polygon
2
B2 Fully correct polygon. Points plotted at the midpoint (B1 All points plotted accurately not joined, or one error in plotting but joined or all points plotted accurately and joined with, additionally, first joined to last or all points at the correct heights and consistently within or at the ends of the intervals and joined (Includes joining last to first to make a polygon)) NB: ignore polygon before 1st point, and after last point. Ignore any histograms.
Mark scheme (b)
Answer
Mark
Notes
\(30 \lt t \leqslant 40\)
1
B1 Allow any notation eg, 30-40 ft polygon
Mark scheme (c)
Working
Answer
Mark
Notes
(6 +2) = 8, (4 + 8 + 14 + 16 + 6 + 2) = 50
\(\dfrac{8}{50}\) oe
2
M1 (6 +2)÷ (4 + 8 + 14 + 16 + 6 + 2) or ft figures from polygon or \(\frac{8}{a}\) with \(a \gt 8\) or \(\frac{c}{50}\) with \(c \lt 50\) or 8 and 50 used but notation incorrect (eg. 8:50 , 8 out of 50) A1 \(\frac{8}{50}\) oe (eg. 0.16) or ft figures from polygon
13 Bob asked each of 40 friends how many minutes they took to get to work.
The table shows some information about his results.
Time taken (\(m\) minutes)
Frequency
\(0 \lt m \leqslant 10\)
3
\(10 \lt m \leqslant 20\)
8
\(20 \lt m \leqslant 30\)
11
\(30 \lt m \leqslant 40\)
9
\(40 \lt m \leqslant 50\)
9
Work out an estimate for the mean time taken. (4)
Mark scheme
Working
Answer
Mark
Notes
5×3+15×8+25×11+35×9+45×9 =1130 1130 ÷ 40
28.25
4
M1 for finding \(fx\) with \(x\) consistent within intervals (including the end points) allow 1 error M1 (dep) for use of all correct mid-interval values M1 (dep on first M1) for \(\Sigma fx \div 40\) or \(\Sigma fx \div \Sigma f\) A1 for 28.25 or \(28\frac{1}{4}\)
12 Rob records the time he takes to drive to work every day for 80 days.
The table shows information about the results.
Time, \(t\) (minutes)
Frequency
\(20 \leqslant t \lt 25\)
16
\(25 \leqslant t \lt 30\)
32
\(30 \leqslant t \lt 40\)
24
\(40 \leqslant t \lt 60\)
8
Total = 80
Last year, the mean time Rob took to drive to work was 25 minutes.
Estimate the percentage increase in the mean driving time for these 80 days. [4 marks]
Mark scheme
Answer
Mark
Comments
\(22.5 \times 16\) or 360 and \(27.5 \times 32\) or 880 and \(35 \times 24\) or 840 and \(50 \times 8\) or 400 or 2480
M1
allow one incorrect midpoint
(their 360 + their 880 + their 840 + their 400) \(\div\) 80 or \(2480 \div 80\) or 31
M1
condone bracket error seen eg \(360 + 880 + 840 + 400 \div 80\) their values must come from ‘midpoints’ within or on class bounds multiplied by correct frequencies
(their \(31 - 25) \div 25\ (\times 100)\) or \(6 \div 25\ (\times 100)\) or \(0.24\ (\times 100)\) or their \(31 \div 25\ (\times 100)\) or \(1.24\ (\times 100)\) or 124
Information about their hourly rates of pay is shown in the table.
Hourly rate, £\(p\)
Number of employees
\(10 \leqslant p \lt 14\)
66
\(14 \leqslant p \lt 20\)
32
\(20 \leqslant p \lt 40\)
15
\(40 \leqslant p \lt 100\)
10
Total \(=\) 123
The owner of the company uses the data to make two statements.
Statement A “Over 30% of employees have an hourly rate that is more than £17”
Statement B “The average hourly rate of pay is more than £20”
(a) Show working that supports Statement A. [3 marks]
(b) Why might Statement A not be true? [1 mark]
(c) Work out an estimate of the mean to support Statement B. [3 marks]
(d) Why is the mean not the best average to represent the data? [1 mark]
Mark scheme (a)
Answer
Mark
Comments
Alternative method 1 Works out best estimate of the percentage of employees with hourly rate more than £17
\(32 \div 2\) or 16
M1
oe implied by 41 or 82
\((15 + 10 + \text{their } 16) \div 123\) or \(41 \div 123\) or \(\dfrac{1}{3}\) or 0.33(...) or \((66 + \text{their } 16) \div 123\) or \(82 \div 123\) or \(\dfrac{2}{3}\) or 0.66(…) or 0.67
23 61 students recorded how many hours they spent revising for a test.
The histogram represents the results.
(a) Work out an estimate of the mean time the 61 students spent revising.
You may use the table to help you. [4 marks]
Time, \(x\) (hours)
Frequency
Midpoint
\(0 \leqslant x \lt 6\)
\(6 \leqslant x \lt 10\)
\(10 \leqslant x \lt 12\)
\(12 \leqslant x \lt 16\)
\(16 \leqslant x \lt 20\)
(b) Give a reason why the answer to part (a) is an estimate. [1 mark]
Mark scheme (a)
Answer
Mark
Comments
\(1.5 \times 6\) or 9 or \(3.5 \times 4\) or 14 or \(5 \times 2\) or 10 or \(4.5 \times 4\) or 18 or \(2.5 \times 4\) or 10
M1
oe values 9, 14, 10 or 18 must be in the correct row in the table or linked to the correct bar on the histogram
\(1.5 \times 6 \times 3\) or \(9 \times 3\) or 27 or \(3.5 \times 4 \times 8\) or \(14 \times 8\) or 112 or \(5 \times 2 \times 11\) or \(10 \times 11\) or 110 or \(4.5 \times 4 \times 14\) or \(18 \times 14\) or 252 or \(2.5 \times 4 \times 18\) or \(10 \times 18\) or 180 or 681
M1dep
oe values 27, 112, 110, 252 or 180 must be in the correct row in the table
(their 27 + their 112 + their 110 + their 252 + their 180) ÷ (their 9 + their 14 + their 10 + their 18 + their 10) or \(681 \div 61\)
M1dep
oe full correct method eg (their 27 + their 112 + their 110 + their 252 + their 180) \(\div\) 61
[11.16, 11.2]
A1
accept 11 with M3 scored and no errors
Additional guidance
Up to M2 may be awarded for correct work with no answer, or incorrect answer, even if this is seen amongst multiple attempts
20 Here is some information about the time spent on social media by 40 women and 40 men last week.
Time spent, \(t\) (hours)
Number of women
Number of men
\(2 \lt t \leqslant 5\)
12
10
\(5 \lt t \leqslant 8\)
11
17
\(8 \lt t \leqslant 11\)
14
9
\(11 \lt t \leqslant 14\)
2
4
\(14 \lt t \leqslant 17\)
1
0
Tick one box for each statement. [3 marks]
Definitely true
Might be true
Cannot be true
Three of the women spent more than 11 hours on social media.
The range for the men is 15 hours.
The women have a higher median than the men.
Mark scheme
Answer
Mark
Comments
Three of the women spent more than 11 hours on social media: Definitely true The range for the men is 15 hours: Cannot be true The women have a higher median than the men: Might be true
B3
B1 for each any clear indication
Additional guidance
Only a cross in a row, mark the cross
A tick and cross(es) in a row – mark the tick
More than one tick in a row scores B0 for that row
16 The number of goals scored by 20 players in a season is shown.
Number of goals
Frequency
Midpoint
0 to 4
6
5 to 9
11
10 to 14
3
Total = 20
Work out an estimate of the mean number of goals per player.
Give your answer as a decimal. [3 marks]
Mark scheme
Answer
Mark
Comments
\(2 \times 6\) or 12 and \(7 \times 11\) or 77 and \(12 \times 3\) or 36 or 125
M1
may be seen in table at least two correct products or their values
\(\dfrac{\text{their } 12 + \text{their } 77 + \text{their } 36}{20}\) or \(\dfrac{125}{20}\) or \(125 \div 20\) or \(6\dfrac{1}{4}\)
M1dep
oe condone bracket error if working seen eg condone \(12 + 77 + 36 \div 20\)
6.25
A1
Additional guidance
6.25 in working, 6 on answer line
M1M1A0
\(125 \div 3\)
M1M0A0
Correct product(s) seen in the table but a different method not using their product(s) used for the mean is choice eg 125 in table but mean calculated as \(20 \div 3 = 6.7\)
13 The amounts spent on clothes by 40 boys and 40 girls in one month were recorded.
The table shows information about the amounts spent by the boys.
Amount, \(x\) (£)
Midpoint
Number of boys
\(0 \leqslant x < 20\)
22
\(20 \leqslant x < 40\)
9
\(40 \leqslant x < 60\)
6
\(60 \leqslant x < 80\)
3
Total = 40
The mean for the girls was £35
Estimate the mean for the girls as a percentage of the mean for the boys. [5 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
Any three of [9.5, 10.5] \(\times\) 22 or [209, 231] and [29.5, 30.5] \(\times\) 9 or [265.5, 274.5] and [49.5, 50.5] \(\times\) 6 or [297, 303] and [69.5, 70.5] \(\times\) 3 or [208.5, 211.5] or 1000
M1
(their [209, 231] + their [265.5, 274.5] + their [297, 303] + their [208.5, 211.5]) \(\div\) 40 or \(1000 \div 40\)
M1dep
oe condone bracket error if working seen eg \(220 + 270 + 300 + 210 \div 40\)
25
A1
\(\dfrac{35}{\text{their } 25}\) or \(\dfrac{7}{5}\) or 1.4
Any three of [9.5, 10.5] \(\times\) 22 or [209, 231] and [29.5, 30.5] \(\times\) 9 or [265.5, 274.5] and [49.5, 50.5] \(\times\) 6 or [297, 303] and [69.5, 70.5] \(\times\) 3 or [208.5, 211.5] or 1000
M1
\(35 \times 40\) or 1400
M1
1000 and 1400
A1
\(\dfrac{\text{their } 1400}{\text{their } 1000}\) or \(\dfrac{7}{5}\) or 1.4
M1dep
oe eg \(1 + \dfrac{\text{their } 1400 - \text{their } 1000}{\text{their } 1000}\) dep on M2
140
A1ft
ft their 1400 and their 1000 with M3 scored
Additional guidance
Alt 1 Correct products seen in the table but a different method not using their products used for the mean shown in the working lines eg \(40 \div 4 = 10\) can score a maximum of M0M0A0M1A1ft
Alt 1 \(1000 \div 4\) (= 250) is not a misread
NB The dependency of the M marks and the requirement for applying A1ft are different for the two alternative methods
Alt 1 3rd M1 Allow any number for their 25 (unless it contradicts their mean)
Alt 1 3rd M1 and A1ft If there is a mean for the boys allow the M mark to be implied by a correct ft answer eg from a mean of 250 allow M1A1ft for 14%
For A1ft allow answers to the nearest whole number or better
Further work after working out the percentage is 3rd M0 eg Mean = 25 \(\dfrac{35}{\text{their } 25} \times 100 = 140\) 140 – 100 = 40 Answer 40
accept 4.8 or 5 if full working shown using correct midpoints
Additional guidance
Two correct from 30, 52.5 and 12.5 implies the first mark and could be used to score up to M2
M1
Midpoints used in the ranges [2, 3], [7, 8] and [12, 13] must be seen eg \(2.5 \times 12\) and \(7 \times 7\) and 12 \((\times 1)\) or \(3 \times 12\) and \(7 \times 7\) and 13 \((\times 1)\) NB These could be used to score up to M2
M1
Correct products seen in the table but a different method shown in the working lines eg \(20 \div 4 = 5\)
\(t \geqslant 15\) product must be 0 if seen condone bracket error seen eg \(30 + 52.5 + 12.5 \div 20\)
4.75
A1
accept 4.8 or 5 if full working shown using correct midpoints
Additional guidance
Two correct from 30, 52.5 and 12.5 implies the first mark and could be used to score up to M2
M1
Midpoints used in the ranges [2, 3], [7, 8] and [12, 13] must be seen eg \(2.5 \times 12\) and \(7 \times 7\) and 12 \((\times 1)\) or \(3 \times 12\) and \(7 \times 7\) and 13 \((\times 1)\) NB These could be used to score up to M2
M1
Correct products seen in the table but a different method shown in the working lines eg \(20 \div 4 = 5\)
23 A farmer has 60 pear trees. The table shows the heights, \(h\) metres, of the pear trees.
Height (\(h\) metres)
Frequency
\(1 < h \leqslant 2\)
5
\(2 < h \leqslant 3\)
8
\(3 < h \leqslant 4\)
32
\(4 < h \leqslant 5\)
15
(a) Calculate an estimate of the mean height of the 60 pear trees. [4]
(b) Explain why it is not possible to use the information from this table to calculate the exact value of the mean height. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
3.45
4
B1 for at least three of 1.5, 2.5, 3.5, 4.5
May be implied by three correct products (7.5, 20, 112, 67.5) or [total =] 207
M1 FT for \(\Sigma mf\) where \(m\) is a consistent value within each group \(1.5 \times 5 + 2.5 \times 8 + 3.5 \times 32 + 4.5 \times 15\) soi by 7.5 + 20 + 112 + 67.5 or 207
FT their “midpoints” seen M1 may be implied by Lower: 5+16+96+60 (177) Upper: 10+24+128+75 (237) Allow one error in calculation
M1 FT dep on M1 for their \(207 \div 60\)
FT from lower 2.95, upper 3.95
Mark scheme (b)
Answer
Marks
Part marks and guidance
Exact heights are not known oe
1
See appendix Do not accept comments on the method used Do not accept “estimate” unless clarified in comment
Appendix
Question 23b
Response
Mark
Because it does not give us an exact height of a pear tree
Correct as mentions “exact height”
1
We don’t know the exact measurements
Accept measurements for heights
1
The heights are not exactly accurate
BOD
1
The data are not accurate
No as could refer to frequencies
0
We don’t know the value of H
Exact value not mentioned
0
We can only estimate the height, we won’t know the exact height
The first part suggests a comment on part (a)
0
Because the metres are in ranges and not precise
Comment on method
0
Because it’s an estimate between two heights
This is a comment on method
0
We have to use the middle value of height each time
This is a comment on method
0
Because the heights are not given
They are but not exact heights
0
It is not accurate
No mention of exact heights not known
0
Due to it being an estimate so the exact can be far off
This is a comment on part (a) and the method
0
Because the heights are given as \(1 < h \leqslant 2\)
This is a comment on method
0
Because it is an estimate and isn’t an exact number
This is a comment on the answer in (a)
0
Because the frequency isn’t exact, it’s an estimate of the height
Incorrect term. “Because the height isn’t exact” would get the mark
0
There is not enough data to tell the exact value of mean height
4 A farmer has 60 pear trees. The table shows the heights, \(h\) metres, of the pear trees.
Height (\(h\) metres)
Frequency
\(1 \lt h \leqslant 2\)
5
\(2 \lt h \leqslant 3\)
8
\(3 \lt h \leqslant 4\)
32
\(4 \lt h \leqslant 5\)
15
(a) Calculate an estimate of the mean height of the 60 pear trees. [4]
(b) Explain why it is not possible to use the information from this table to calculate the exact value of the mean height. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
3.45
4
B1 for at least three of 1.5, 2.5, 3.5, 4.5
May be implied by three correct products (7.5, 20, 112, 67.5) or [total =] 207
M1 FT for \(\Sigma mf\) where \(m\) is a consistent value within each group 1.5 × 5 + 2.5 × 8 + 3.5 × 32 + 4.5 × 15 soi by 7.5 + 20 + 112 + 67.5 or 207
FTtheir “midpoints” seen M1 may be implied by Lower: 5+16+96+60 (177) Upper: 10+24+128+75 (237) Allow one error in calculation
M1 FT dep on M1 for their 207 ÷ 60
FT from lower 2.95, upper 3.95
Mark scheme (b)
Answer
Marks
Part marks and guidance
Exact heights are not known oe
1
See appendix Do not accept comments on the method used Do not accept “estimate” unless clarified in comment
Appendix: Exemplar responses for Q4(b)
Response
Mark
Because it does not give us an exact height of a pear tree
Correct as mentions “exact height”
1
We don’t know the exact measurements
Accept measurements for heights
1
The heights are not exactly accurate
BOD
1
The data are not accurate
No as could refer to frequencies
0
We don’t know the value of H
Exact value not mentioned
0
We can only estimate the height, we won’t know the exact height
The first part suggests a comment on part (a)
0
Because the metres are in ranges and not precise
Comment on method
0
Because it’s an estimate between two heights
This is a comment on method
0
We have to use the middle value of height each time
This is a comment on method
0
Because the heights are not given
They are but not exact heights
0
It is not accurate
No mention of exact heights not known
0
Due to it being an estimate so the exact can be far off
This is a comment on part (a) and the method
0
Because the heights are given as \(1 \lt h \leqslant 2\)
This is a comment on method
0
Because it is an estimate and isn’t an exact number
This is a comment on the answer in (a)
0
Because the frequency isn’t exact, it’s an estimate of the height
Incorrect term. “Because the height isn’t exact” would get the mark
0
There is not enough data to tell the exact value of mean height
19 A fitness centre records how long each customer spends in the gym. This cumulative frequency table summarises the results.
Time (t minutes)
Cumulative frequency
\(t \leqslant 10\)
6
\(t \leqslant 20\)
24
\(t \leqslant 30\)
35
\(t \leqslant 40\)
48
\(t \leqslant 50\)
60
\(t \leqslant 60\)
74
Calculate an estimate of the mean time the customers spend in the gym. You must show your working. You may use the table above to help present your work. [5]
Mark scheme
Answer
Marks
Part marks and guidance
31.6[2…] with correct working
5
Accept 32 with correct working Correct working requires at least evidence of B1 B1 M1
B1 for at least 4 correct frequencies from 6, 18, 11, 13, 12, 14
B1 for at least 4 correct midpoints from 5, 15, 25, 35, 45, 55
May be implied by correct products seen 30, 270, 275, 455, 540, 770
May be implied by correct products seen 30, 270, 275, 455, 540, 770
M1 for \(\sum mf\) using their midpoints and their frequencies soi by 30+270+275+455+540+770 or by 2340
their midpoints must lie in the range of t for each interval Lower: [0 or 6]+180+220+390+480+700 Upper: 60+360+330+520+600+840 Allow one error in calculation their frequencies must not be the cumulative frequencies
M1dep for their \(\frac{\sum mf}{74}\)
If 0 or 1 scored, instead award SC2 for final answer 31.6[2…] with no working or insufficient working
If 0 scored, instead award SC1 for 2340 with no working or insufficient working
(a) The masses, \(m\) kg, of some parcels are shown below.
4 15 14 11 12 3 1 18 13 2 16 10
Jack constructs this grouped frequency table to record the masses.
Mass (m kg)
Tally
Frequency
\(0 \leqslant m \leqslant 5\)
\(5 \leqslant m \leqslant 10\)
\(10 \leqslant m \leqslant 15\)
\(15 \leqslant m \leqslant 20\)
Explain why Jack’s table is unsuitable to record the masses. [1]
(b) The histogram summarises the times taken, in minutes, by some students to complete a race.
(i) Show that 70 students took between 45 and 70 minutes to complete the race. [2]
(ii) Calculate an estimate of the mean time taken to complete the race. Show your working. [5]
Mark scheme (a)
Answer
Marks
Part marks and guidance
Refers to overlapping intervals
1
eg 10 could go into 2 intervals The same number can go in 2 places Upper value in interval should be < Both inequalities are ≤ when only one should be
Mark scheme (b)
Answer
Marks
Part marks and guidance
(i) 5 × 6 and 2 × 20
2
M1 for 5 × 6 or 2 × 20
Could be written on graph Allow eg 2 × 10 + 2 × 10 for 2 × 20 Not just 30 + 40, must show products
(ii) 50.25 with correct working
5
B1 for frequencies 10, 20, 30, 40 M1 for mid-interval values 35, 42.5, 47.5, 60 soi M1 for \(\sum ft\) where \(t\) is in the interval including boundaries FT their frequencies
M1 for \(\sum ft \div \sum f\) dep on previous M1 FT their frequencies
If 0 scored, SC2 for answer 50.25 or SC1 for 5025 with no working
“Correct working” requires evidence of at least B1M1M1 Condone 1 error, could be on graph,
12 The cumulative frequency graph summarises the annual salary, \(p\) (£ thousands), of the 60 workers in a factory.
(a) Use the graph to estimate the median annual salary. [1]
(b) Complete this cumulative frequency table.
Annual salary, p (£ thousands)
Cumulative frequency
\(p \leqslant 10\)
\(p \leqslant 20\)
\(p \leqslant 30\)
\(p \leqslant 50\)
\(p \leqslant 80\)
[2]
(c) Use the information in the cumulative frequency table to calculate an estimate of the mean annual salary. [5]
(d) Explain why your estimate of the median is more reliable than your estimate of the mean. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
24
1
condone 24 000
Mark scheme (b)
Answer
Marks
Part marks and guidance
14 26 36 50 60
2
B1 for complete table with three correct
Mark scheme (c)
Answer
Marks
Part marks and guidance
28.5
5
B1FT for frequencies 14, 12, 10, 14, 10 with one error (allow FT table in (b)). B1 for 5, 15, 25, 40, 65 (allow one error) M1 for \(\sum(\textit{their midpoint} \times \textit{their frequency})\)
M1 for their 1710 ÷ 60
implied by 70 + 180 + 250 + 560 + 650 or 1710 their midpoint must be within the group range
Mark scheme (d)
Answer
Marks
Part marks and guidance
Acceptable answer e.g. for the mean the figures used are approximate
1
See exemplars in appendix
Appendix: Exemplar responses for Q12(d)
Response
Mark
Because the median estimate is only affected by mistakes in measuring
1
Because the mean relies on assumption that all values lie at midpoint
1
Median is accurately read from graph, mean is estimated ….
1
Number of people earning each amount is not evenly spread out …..
20 Luke is an office receptionist. Each day, for 60 days, he records the number of people visiting the office.
Number of people, (\(n\))
Frequency
\(0 \leqslant n \leqslant 5\)
20
\(5 \lt n \leqslant 10\)
14
\(10 \lt n \leqslant 20\)
11
\(20 \lt n \leqslant 40\)
15
(a) Calculate an estimate of the mean number of people visiting the office. [4]
(b) Luke says the range is 40.
Explain why he may be wrong. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
12.8[3....]
4
B1 for at least 3 mid-points seen (from 2.5, 7.5, 15, 30) or implied by products 50,105,165,450 or 770
M1 for \(\Sigma mf\) where \(m\) is a value within each group. Allow use of boundaries; allow one error in calculation. If no midpoints seen may be implied by their mf
M1 dep on previous M1 for their \(770 \div 60\)
Mark scheme (b)
Answer
Marks
Part marks and guidance
The highest number may not have been 40 or the lowest number may not have been 0.
1
See appendix
Appendix
Exemplar responses for Q20(b)
Response
Mark
Because you don’t know the exact biggest and smallest numbers of people
1
Exact number of people is not known, they are just boundaries where they could be
1
No exact numbers given/don’t have exact values
1
40 – 5 = 35 so he could be wrong
1
You don’t know the exact weight
0
Not exactly 40 people visited the office
0
Because we are using midpoints so won’t get accurate result
17 The police record the speed of vehicles passing a speed checkpoint. The speeds are recorded in the table below.
Speed (\(s\) mph)
Number of vehicles
\(0 \lt s \le 20\)
5
\(20 \lt s \le 40\)
8
\(40 \lt s \le 50\)
37
\(50 \lt s \le 60\)
47
\(60 \lt s \le 80\)
3
(a) Calculate an estimate of the mean speed of the vehicles. [4]
(b) Explain why it is not possible to use the information from this table to calculate the exact value of the mean speed. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
47.5
4
B1 for at least four of 10, 30, 45, 55, 70
M1 FT for \(\Sigma mf\) where \(m\) is a value within each group 10\(\times\)5 + 30\(\times\)8 + 45\(\times\)37 + 55\(\times\)47 + 70\(\times\)3 soi by 50+240+1665+2585+210 or 4750
M1 FT dep on M1 for their 4750 \(\div\) their (5+8+37+47+3)
May be implied by four correct products or 4750
FT their “midpoints” seen. M1 may be implied by Lower: 0+160+1480+2350+180 (4170) (corrected from the printed mark scheme: 4180) Upper: 100+320+1850+2820+240 (5330) Allow one error in calculation
Expect 100
Mark scheme (b)
Answer
Marks
Part marks and guidance
Exact speeds for each vehicle are not recorded oe
1
Do not accept, “Because the mid-point is used” or comments on the method used. Accept e.g.: Specific speeds not given or We don’t know the speeds The exact speed isn’t given
2 The police record the speed of vehicles passing a speed checkpoint. The speeds are recorded in the table below.
Speed (s mph)
Number of vehicles
\(0 \lt s \leqslant 20\)
5
\(20 \lt s \leqslant 40\)
8
\(40 \lt s \leqslant 50\)
37
\(50 \lt s \leqslant 60\)
47
\(60 \lt s \leqslant 80\)
3
(a) Calculate an estimate of the mean speed of the vehicles. [4]
(b) Explain why it is not possible to use the information from this table to calculate the exact value of the mean speed. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
47.5
4
B1 for at least four of 10, 30, 45, 55, 70 M1 FT for Σmf where m is a value within each group 10×5 + 30×8 + 45×37 + 55×47 + 70×3 soi by 50 + 240 + 1665 + 2585 + 210 or 4750 M1 FT dep on M1 for their 4750 ÷ their (5 + 8 + 37 + 47 + 3)
May be implied by four correct products or 4750
FT their “midpoints” seen. M1 may be implied by Lower: 0 + 160 + 1480 + 2350 + 180 (4180) Upper: 100 + 320 + 1850 + 2820 + 240 (5330) Allow one error in calculation.
Expect 100
Mark scheme (b)
Answer
Marks
Part marks and guidance
Exact speeds for each vehicle are not recorded oe
1
Do not accept, “Because the mid-point is used” or comments on the method used. Accept e.g.: Specific speeds not given or We don’t know the speeds The exact speed isn’t given
