13 Alan grew 80 plants of the same type outside. The cumulative frequency graph shows information about the heights, in cm, of these plants.
One of the plants is chosen at random.
(a) Find an estimate for the probability that this plant will have a height greater than 90 cm. (2)
(b) Use the graph to find an estimate for the median height. (1)
(c) Use the graph to find an estimate for the interquartile range of the heights. (2)
Alan also grew plants of the same type inside. The interquartile range of the heights of these plants is 30 cm.
(d) Give one comparison between the distribution of the heights of the plants grown inside with the distribution of the heights of the plants grown outside. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
\(\dfrac{6}{80}\)
M1
for correct method using the graph at 90 cm, eg \(80 - 74\ (= 6)\)
A1
for \(\dfrac{6}{80}\) oe, accept \(\dfrac{5}{80}\) oe
Additional guidance
Accept \(80 - 75\ (= 5)\)
Accept any equivalent fraction, decimal (0.075) or percentage (7.5%)
Mark scheme (b)
Answer
Mark
Mark scheme
60
B1
for answer in the range 58.5 – 61.5
Mark scheme (c)
Answer
Mark
Mark scheme
35
M1
for UQ = 75 or LQ = 40
A1
for answers in the range 34 to 36 or ft acceptable readings from the graph
Additional guidance
Allow UQ in range 74 to 76 Allow LQ in range 39 to 41 Value(s) may be seen on graph
Mark scheme (d)
Answer
Mark
Mark scheme
Comparison made
C1
(ft) for a correct comparison of the IQRs eg the heights of the plants grown outside are more varied or the IQR of the plants grown outside is 35 cm but the IQR of the plants grown inside is only 30 cm.
Additional guidance
The comparison must include some context (eg refers to heights or cm) Ignore any additional information provided there is no contradiction. Figures need not be seen but if given they must be correct.
for 4 or 5 points plotted correctly from the cumulative frequency table
A1
for a fully correct graph SCB1 if 4 or 5 points plotted not at end but consistent within each interval and joined by a curve or line segments providing no gradient is negative
Additional guidance
Ignore to the left of the first point and right of the last point If histograms drawn, points must be identified
12 The cumulative frequency table gives information about the heights, in cm, of 40 plants.
Height (\(h\) cm)
Cumulative Frequency
\(0 \lt h \leqslant 5\)
4
\(0 \lt h \leqslant 10\)
11
\(0 \lt h \leqslant 15\)
24
\(0 \lt h \leqslant 20\)
34
\(0 \lt h \leqslant 25\)
38
\(0 \lt h \leqslant 30\)
40
(a) On the grid below, draw a cumulative frequency graph for this information.(2)
(b) Use the graph to find an estimate for the median height of the plants. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
cf graph
M1
for 5 or 6 points plotted correctly
A1
for a fully correct graph
SC B1 if 5 or 6 of their points plotted not at the end but consistent within each interval and joined by a curve or line segments providing no gradient is negative
Additional guidance
If histograms drawn, points must be identified
Accept a smooth curve or line segments Ignore to the left of the first point and right of the last point
Mark scheme (b)
Answer
Mark
Mark scheme
13 to 14
B1
for answer in the range 13 to 14 or ft their cf graph
8 The grouped frequency table gives information about the time, in minutes, taken by 50 people to solve a puzzle.
Time (\(t\) minutes)
Frequency
\(0 \lt t \leqslant 10\)
5
\(10 \lt t \leqslant 20\)
8
\(20 \lt t \leqslant 30\)
12
\(30 \lt t \leqslant 40\)
15
\(40 \lt t \leqslant 50\)
7
\(50 \lt t \leqslant 60\)
3
Brian was asked to draw a cumulative frequency table for this information.
This is the table that Brian drew.
Time (\(t\) minutes)
Cumulative frequency
\(0 \lt t \leqslant 10\)
5
\(10 \lt t \leqslant 20\)
13
\(20 \lt t \leqslant 30\)
25
\(30 \lt t \leqslant 40\)
40
\(40 \lt t \leqslant 50\)
47
\(50 \lt t \leqslant 60\)
50
Write down one thing that is wrong with this cumulative frequency table. (1)
Mark scheme
Answer
Mark
Mark scheme
Error in inequalites
C1
for identifying incorrect inequalities Acceptable examples gives at least one correct inequality eg \((10 \lt t \leqslant 20)\) should be \(0 \lt t \leqslant 20\) it should be \(t \leqslant 20\) (all) inequalities should start with 0 should start with 0
Not acceptable examples \(10 \lt t \leqslant 20\) is wrong the numbers have been added wrong
10 The cumulative frequency table shows information about the times, in minutes, taken by 40 people to complete a puzzle.
Time (\(m\) minutes)
Cumulative frequency
\(20 \lt m \leqslant 40\)
5
\(20 \lt m \leqslant 60\)
25
\(20 \lt m \leqslant 80\)
35
\(20 \lt m \leqslant 100\)
38
\(20 \lt m \leqslant 120\)
40
(a) On the grid below, draw a cumulative frequency graph for this information.(2)
(b) Use your graph to find an estimate for the interquartile range. (2)
One of the 40 people is chosen at random.
(c) Use your graph to find an estimate for the probability that this person took between 50 minutes and 90 minutes to complete the puzzle. (2)
Mark scheme (a)
Answer
Mark
Mark scheme
cf graph through (40, 5), (60, 25), (80, 35), (100, 38) and (120, 40)
C2
for a complete and accurate cf graph
(C1
for at least 4 or 5 cf values plotted correctly)
SC: B1 for 4 or 5 points plotted not at end but consistently within each interval and joined provided no gradient is negative
Additional guidance
May be a cumulative frequency curve or a cumulative frequency polygon Ignore any graph drawn to the left of the first point If histograms drawn, plots must be identified
Mark scheme (b)
Answer
Mark
Mark scheme
answer in range 21 to 28
M1
for UQ in the range 66 to 70 or LQ in the range 42 to 46 or ft their cf graph
A1
for answer in range 21 to 28 or ft their cf graph
Mark scheme (c)
Answer
Mark
Mark scheme
answer in the range \(\dfrac{19}{40}\) to \(\dfrac{24}{40}\)
M1
for finding the difference between readings taken from the cf axis at points from a mark of 50 and a mark of 90 or ft their graph (if possible)
A1
for an answer in the range \(\dfrac{19}{40}\) to \(\dfrac{24}{40}\) or ft their cf graph
Additional guidance
Their graph must be a cf graph
Accept any equivalent fraction, decimal from 0.475 to 0.6 or percentage from 47.5% – 60%
11 The grouped frequency table gives information about the times, in minutes, that 80 office workers take to get to work.
Time (\(t\) minutes)
Frequency
\(0 \lt t \leqslant 20\)
5
\(20 \lt t \leqslant 40\)
30
\(40 \lt t \leqslant 60\)
20
\(60 \lt t \leqslant 80\)
15
\(80 \lt t \leqslant 100\)
8
\(100 \lt t \leqslant 120\)
2
(a) Complete the cumulative frequency table.
Time (\(t\) minutes)
Cumulative frequency
\(0 \lt t \leqslant 20\)
\(0 \lt t \leqslant 40\)
\(0 \lt t \leqslant 60\)
\(0 \lt t \leqslant 80\)
\(0 \lt t \leqslant 100\)
\(0 \lt t \leqslant 120\)
(1)
(b) On the grid, draw the cumulative frequency graph for this information.(2)
(c) Use your graph to find an estimate for the percentage of these office workers who take more than 90 minutes to get to work. (3)
Mark scheme (a)
Answer
Mark
Mark scheme
5, 35, 55, 70, 78, 80
B1
cao
Mark scheme (b)
Answer
Mark
Mark scheme
cf graph
M1
for 5 or 6 of their points plotted correctly from a cf table
A1
for a fully correct graph
SC B1 if 5 or 6 of their points plotted not at end but consistent within each interval and joined by a curve or line segments providing no gradient is negative
Additional guidance
Ignore to the left of the first point and right of the last point
Accept a smooth curve or line segments
Mark scheme (c)
Answer
Mark
Mark scheme
7.5
M1
for a clear method to read off the cf graph at 90
M1
for a full method to find the percentage eg \((80 - \text{``}74\text{''}) \div 80 \times 100\ (= 7.5)\)
A1
for 7.5 or ft cf graph
Additional guidance
Sight of 74 or 6 implies M1
The following readings give the following percentages 72 = 10% 73 = 8.75% 74 = 7.5% 75 = 6.25% 76 = 5%
9 The times that 48 trains left a station on Monday were recorded.
The cumulative frequency graph gives information about the numbers of minutes the trains were delayed, correct to the nearest minute.
The shortest delay was 0 minutes. The longest delay was 42 minutes.
(a) On the grid below, draw a box plot for the information about the delays on Monday.(3)
48 trains left the station on Tuesday. The box plot below gives information about the delays on Tuesday.
(b) Compare the distribution of the delays on Monday with the distribution of the delays on Tuesday. (2)
Mary says,
“The longest delay on Tuesday was 33 minutes. This means that there must be some delays of between 25 minutes and 30 minutes.”
(c) Is Mary right? You must give a reason for your answer. (1)
Mark scheme (a)
Answer
Mark
Mark scheme
box plot drawn
B1
ends of whiskers at 0 and 42 with a box
B1
median at 10 inside a box
B1
for ends of box at 4 and 20
Additional guidance
The box can be of any height. Accept ends that are marked (eg line, cross, dot) or defined by the end of the whiskers if clear.
Has to be inside a box; whiskers not required
An independent mark that can be awarded for just a box; do not need whiskers for this mark.
Mark scheme (b)
Answer
Mark
Mark scheme
Comparison
C1
for a correct comparison of medians, eg. the median delay time on Mon was greater than the median delay time on Tues. or ft (a)
C1
for a correct comparison of a measure of spread, eg. the interquartile range (range) of delay times on Mon was greater than the interquartile range (range) of delay times on Tues. or ft (a) For the award of both marks at least one of the comparisons must be in context
Additional guidance
Simply quoting values for median, range and IQR is insufficient, they must be compared
Comparisons can relate to the median, and then either the range or the IQR.
Mark scheme (c)
Answer
Mark
Mark scheme
statement
C1
‘No’ with statement explaining that there might not be any delays between 25 minutes and 30 minutes as in the upper 25% (12 trains) the delays may all be between 17 and 25 or 30 and 33
Additional guidance
The ‘No’ may be implied from their wording, and could be written next to the “?” The statement must mention (or imply) values above the UQ of 17
16 The table gives information about the marks gained by some students in an exam.
Mark (\(m\))
Frequency
\(0 \lt m \leqslant 20\)
40
\(20 \lt m \leqslant 40\)
70
\(40 \lt m \leqslant 60\)
60
\(60 \lt m \leqslant 80\)
15
\(80 \lt m \leqslant 100\)
10
\(100 \lt m \leqslant 120\)
5
(a) Complete the cumulative frequency table for this information.
Mark (\(m\))
Cumulative frequency
\(0 \lt m \leqslant 20\)
\(0 \lt m \leqslant 40\)
\(0 \lt m \leqslant 60\)
\(0 \lt m \leqslant 80\)
\(0 \lt m \leqslant 100\)
\(0 \lt m \leqslant 120\)
(1)
(b) On the grid, draw a cumulative frequency graph for your table.(2)
(c) Use your graph to find an estimate for the number of students who gained a mark of more than 54 (2)
Mark scheme (a)
Working
Answer
Mark
Notes
40, 110, 170, 185, 195, 200
Table
1
B1
Mark scheme (b)
Answer
Mark
Notes
Cumulative frequency diagram
2
M1 ft their cumulative frequency table for at least 5 points plotted correctly at the ends of the intervals provided tables values are cumulative, condoning one arithmetical error, or if the shape of the graph is correct for 5 or 6 points plotted not at the ends but consistently within each interval and joined. A1 for a correct graph (allow curve or line segments)
Mark scheme (c)
Answer
Mark
Notes
40 to 48
2
M1 for reading their cumulative frequency graph from mark of 54 (= 152 to 160) where the points are plotted consistently within each interval and joined. A1 for answer in the range 40 to 48 or ft from their cumulative frequency graph
13 Charlotte grows some potatoes. The table shows information about the weights of her potatoes.
Weight (\(w\) grams)
Frequency
\(100 \lt w \leqslant 120\)
5
\(120 \lt w \leqslant 140\)
25
\(140 \lt w \leqslant 160\)
30
\(160 \lt w \leqslant 180\)
15
\(180 \lt w \leqslant 200\)
5
(a) Complete the cumulative frequency table.
Weight (\(w\) grams)
Cumulative frequency
\(100 \lt w \leqslant 120\)
\(100 \lt w \leqslant 140\)
\(100 \lt w \leqslant 160\)
\(100 \lt w \leqslant 180\)
\(100 \lt w \leqslant 200\)
(1)
(b) On the grid opposite, draw a cumulative frequency graph for your table.(2)
(c) Use your graph to find an estimate for the interquartile range. (2)
(d) Use your graph to find an estimate for the percentage of Charlotte’s potatoes with a weight less than 150 grams. (3)
Mark scheme (a)
Answer
Mark
Notes
5, 30, 60, 75, 80
1
B1 for correct cumulative frequencies (may be implied by correct heights on the grid)
Mark scheme (b)
Answer
Mark
Notes
cf graph
2
M1 for at least 4 of the 5 points plotted correctly at the ends of the intervals or 4 of the 5 points plotted not at the ends but consistently within each interval and joined (dep on a cf table with no more than one arithmetic error) A1 for a fully correct cf graph (points may be joined by a curve or straight line segments)
Mark scheme (c)
Working
Answer
Mark
Notes
IQR = UQ – LQ
26-28
2
M1 for reading values from their cf graph at cf = 20 or 20.25 and cf = 60 or 60.75 A1ft provided M1 is awarded in (b)
Mark scheme (d)
Answer
Mark
Notes
55-59
3
M1 for reading a value from their cf graph at weight 150 grams M1 for \(\dfrac{\text{``}45\text{''}}{\text{``}80\text{''}} \times 100\) A1ft provided M1 is awarded in (b)
One day the company delivered 80 parcels. The table shows information about the weights, in kg, of these parcels.
Weight (\(w\) kg)
Frequency
\(0 \lt w \leqslant 1\)
19
\(1 \lt w \leqslant 2\)
17
\(2 \lt w \leqslant 3\)
15
\(3 \lt w \leqslant 4\)
12
\(4 \lt w \leqslant 5\)
10
\(5 \lt w \leqslant 6\)
7
(a) Complete the cumulative frequency table.
Weight (\(w\) kg)
Cumulative frequency
\(0 \lt w \leqslant 1\)
\(0 \lt w \leqslant 2\)
\(0 \lt w \leqslant 3\)
\(0 \lt w \leqslant 4\)
\(0 \lt w \leqslant 5\)
\(0 \lt w \leqslant 6\)
(1)
(b) On the grid, draw a cumulative frequency graph for your table.(2)
Sue says,
“75% of the parcels weigh less than 3.4 kg.”
*(c) Is Sue correct? You must show how you get your answer. (3)
Mark scheme (a)
Answer
Mark
Notes
19, 36, 51, 63, 73, 80
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
cf graph
2
M1 for at least 5 of the 6 points plotted at each upper end of the interval (not joined) or 5 of the 6 points plotted consistently within interval (not upper end) and joined (dep on a cf table with no more than one arithmetic error) A1 correct graph
Mark scheme (c)
Answer
Mark
Notes
comparable value and conclusion
3
M1 for indication of a reading taken from a cf graph using weight = 3.4 kg or find UQ from 60 A1 for value given between 55 & 57 or 3.6 & 3.8 C1 (dep on at least M1) for conclusion (justified)
16 The table shows information about the times taken by 100 people in a fun run.
Time (\(t\) minutes)
Frequency
\(20 \lt t \leqslant 30\)
4
\(30 \lt t \leqslant 40\)
16
\(40 \lt t \leqslant 50\)
36
\(50 \lt t \leqslant 60\)
24
\(60 \lt t \leqslant 70\)
14
\(70 \lt t \leqslant 80\)
6
(a) Complete the cumulative frequency table for this information.
Time (\(t\) minutes)
Cumulative frequency
\(20 \lt t \leqslant 30\)
\(20 \lt t \leqslant 40\)
\(20 \lt t \leqslant 50\)
\(20 \lt t \leqslant 60\)
\(20 \lt t \leqslant 70\)
\(20 \lt t \leqslant 80\)
(1)
(b) On the grid, draw a cumulative frequency graph for your table.(2)
(c) Use your graph to find an estimate for the median time. (1)
(d) Use your graph to find an estimate for the number of people who took longer than 63 minutes. (2)
Mark scheme (a)
Answer
Mark
Notes
4, 20, 56, 80, 94, 100
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
graph
2
M1 ft from their table for at least 5 points plotted correctly at the ends of the intervals provided table values are cumulative, condoning one arithmetic error, or if the shape of the graph is correct for 5 or 6 points plotted not at the ends but consistently within each interval and joined A1 cao for correct graph with points joined by curve or straight line segments
Mark scheme (c)
Answer
Mark
Notes
47 to 49
1
B1 for 47 to 49 or ft their cf graph at cf = 50
Mark scheme (d)
Answer
Mark
Notes
13 to 16
2
M1 for reading a value from their cf graph at time = 63 (84 to 87) A1 for answer in the range 13 to 16 or ft from their graph
16 There are 200 workers at a factory. The cumulative frequency table gives information about their ages.
Age (\(a\) years)
Cumulative frequency
\(0 \lt a \leqslant 20\)
25
\(0 \lt a \leqslant 30\)
70
\(0 \lt a \leqslant 40\)
138
\(0 \lt a \leqslant 50\)
175
\(0 \lt a \leqslant 60\)
186
\(0 \lt a \leqslant 70\)
194
\(0 \lt a \leqslant 80\)
200
(a) On the grid, draw a cumulative frequency graph for this information.(2)
(b) Graham says, “10% of workers at the factory are older than 65” Is Graham correct? You must show how you get your answer. (2)
Mark scheme (a)
Answer
Mark
Notes
correct graph
2
M1 for 5 or 6 or 7 points plotted correctly at the ends of the intervals (overlay) A1 cao for correct graph with points joined by curve or straight line segments [SC: B1 if the shape of the graph is correct and 5 or 6 or 7 of their points are not at the ends but are plotted consistently within (10,20) (20,30) (30,40) etc.]
Mark scheme (b)
Answer
Mark
Notes
No with supporting figures
2
M1 for \(0.1 \times 200\) (=20) or \(0.9 \times 200\) (=180) or sight of 180 used on cf axis or \(200 - 186\) (=14) A1 ft for correct decision with 20 and “9” or 20 and 14 or “age” from reading graph at 180
OR M1 for method to find percentage of workers who are over 65, eg \(\dfrac{200 - \text{``}191\text{''}}{200} \times 100\) (=4.5%) or method to find percentage of workers who are over 60 (from table), eg \(\dfrac{200 - 186}{200} \times 100\) (=7%) or \(\dfrac{200 - 190}{200} \times 100\) (=5%) A1 ft for correct decision with “4.5”% or 7% or 5%
18 The cumulative frequency graph shows information about the times 80 swimmers take to swim 50 metres.
(a) Use the graph to find an estimate for the median time. (1)
A swimmer has to swim 50 metres in 60 seconds or less to qualify for the swimming team.
The team captain says,
“More than 25% of swimmers have qualified for the swimming team.”
*(b) Is the team captain right? You must show how you got your answer. (3)
For these 80 swimmers the least time taken was 28 seconds and the greatest time taken was 96 seconds.
(c) Use the cumulative frequency graph and the information above to draw a box plot for the times taken by the swimmers.(3)
Mark scheme (a)
Answer
Mark
Notes
68
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
Yes as \(28 \gt 20\) or \(35\% \gt 25\%\) or \(53 \lt 60\)
3
M1 for reading a value from graph at time = 60 (=28, accept 27 to 28) M1 for ‘28’ \(\div 80 \times 100\) (= 35) or \(25 \div 100 \times 80\) (= 20) C1 (dep on M2) for correct decision based on their figures
OR
M1 for \(25 \div 100 \times 80\) (= 20) M1 for reading a value from graph at cf = 20 (=53, accept 52 to 54) C1 (dep on M2) for correct decision based on their figures
Mark scheme (c)
Working
Answer
Mark
Notes
28, 53, 68, 76, 96
Box plot plotted
3
B1 for ends of whiskers at 28 and 96 with a box B1 ft for median at ‘68’ inside a box B1 for ends of box at 53 (accept 52 to 54) and 76
21 The table below shows information about the heights of 60 students.
Height (\(x\) cm)
Number of students
\(140 \lt x \leqslant 150\)
4
\(150 \lt x \leqslant 160\)
5
\(160 \lt x \leqslant 170\)
16
\(170 \lt x \leqslant 180\)
27
\(180 \lt x \leqslant 190\)
5
\(190 \lt x \leqslant 200\)
3
(a) On the grid opposite, draw a cumulative frequency graph for the information in the table.(3)
(b) Find an estimate
(i) for the median,
(ii) for the interquartile range. (3)
Mark scheme (a)
Working
Answer
Mark
Notes
Cf table: 4, 9, 25, 52, 57, 60
Correct Cf graph
3
B1 Correct cumulative frequencies (may be implied by correct heights on the grid) M1 for at least 5 of “6 points” plotted consistently within each interval A1 for a fully correct CF graph
Mark scheme (b)
Working
Answer
Mark
Notes
(i) 172
B1 for 172 or read off at cf = 30 or 30.5 from a cf graph, ft provided M1 is awarded in (a)
IQR = UQ – LQ
(ii) 12 - 14
3
M1 for readings from graph at cf = 15 or 15.25 and cf = 45 or 45.75 from a cf graph with at least one of LQ or UQ correct from graph (± ½ square). A1ft provided M1 is awarded in (a)
14 The grouped frequency table shows information about the weekly wages of 80 factory workers.
Weekly wage (£\(x\))
Frequency
\(100 \lt x \leqslant 200\)
8
\(200 \lt x \leqslant 300\)
15
\(300 \lt x \leqslant 400\)
30
\(400 \lt x \leqslant 500\)
17
\(500 \lt x \leqslant 600\)
7
\(600 \lt x \leqslant 700\)
3
(a) Complete the cumulative frequency table.
Weekly wage (£\(x\))
Cumulative Frequency
\(100 \lt x \leqslant 200\)
\(100 \lt x \leqslant 300\)
\(100 \lt x \leqslant 400\)
\(100 \lt x \leqslant 500\)
\(100 \lt x \leqslant 600\)
\(100 \lt x \leqslant 700\)
(1)
(b) On the grid opposite, draw a cumulative frequency graph for your table. (2)
(c) Use your graph to find an estimate for the interquartile range. (2)
(d) Use your graph to find an estimate for the number of workers with a weekly wage of more than £530 (2)
Mark scheme (a)
Answer
Mark
Notes
8, 23, 53, 70, 77, 80
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
graph
2
M1 ft from their table for at least 5 points plotted correctly at the ends of the intervals provided table values are cumulative, condoning one arithmetic error A1 cao for correct graph with points joined by curve or straight line segments [SC B1 if the shape of the graph is correct and 5 points of their points are not at the ends but consistently within each interval and joined.]
Mark scheme (c)
Working
Answer
Mark
Notes
Readings at 60 and 20 420 to 440 – 280 to 295
120 – 160
2
M1 (dep on cf graph) for use of either cf = 20 or cf = 60 A1 ft from a cf graph
Mark scheme (d)
Working
Answer
Mark
Notes
80 – 71 to 74
6 – 9
2
M1 (dep on cf graph) for evidence of reading off the cf axis from £530 on the wages axis (could be the answer) A1ft for 6 - 9
21 The table shows information about the speeds of 100 lorries.
Speed (\(s\)) in km/h
Frequency
\(0 \lt s \leqslant 20\)
2
\(20 \lt s \leqslant 40\)
9
\(40 \lt s \leqslant 60\)
23
\(60 \lt s \leqslant 80\)
31
\(80 \lt s \leqslant 100\)
27
\(100 \lt s \leqslant 120\)
8
(a) Complete the cumulative frequency table for this information.
Speed (\(s\)) in km/h
Cumulative frequency
\(0 \lt s \leqslant 20\)
2
\(0 \lt s \leqslant 40\)
\(0 \lt s \leqslant 60\)
\(0 \lt s \leqslant 80\)
\(0 \lt s \leqslant 100\)
\(0 \lt s \leqslant 120\)
(1)
(b) On the grid, draw a cumulative frequency graph for your table.(2)
(c) Find an estimate for the number of lorries with a speed of more than 90 km/h. (2)
Mark scheme (a)
Answer
Mark
Notes
11, 34, 65, 92, 100
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
cf graph
2
B1 for 5 or 6 points plotted correctly ±1 full 2 mm square at the upper end of the interval dep on sensible table (condone one error in addition) B1 (dep) for points joined by curve or line segments provided no gradient is negative. Ignore any point or graph outside range of their points. SC B1 for 5 or 6 points plotted not at end but consistently within each interval and joined.
Mark scheme (c)
Answer
Mark
Notes
18 – 24
2
M1 for indication of taking a reading from 90 or ft from their cf graph A1 for 18 – 24
This year he put his tomato plants into two groups, group A and group B.
Harry gave fertiliser to the tomato plants in group A. He did not give fertiliser to the tomato plants in group B.
Harry weighed 60 tomatoes from group A. The cumulative frequency graph shows some information about these weights.
(a) Use the graph to find an estimate for the median weight. (1)
The 60 tomatoes from group A had a minimum weight of 153 grams and a maximum weight of 186 grams.
(b) Use this information and the cumulative frequency graph to draw a box plot for the 60 tomatoes from group A.(3)
Harry did not give fertiliser to the tomato plants in group B. Harry weighed 60 tomatoes from group B. He drew this box plot for his results.
(c) Compare the distribution of the weights of the tomatoes from group A with the distribution of the weights of the tomatoes from group B. (2)
Mark scheme (a)
Answer
Mark
Notes
170
1
B1 accept answers in range 170 - 170.5 inclusive
Mark scheme (b)
Answer
Mark
Notes
3
B3 for box plot with all 3 aspects correct (overlay) aspect 1 : ends of whiskers at 153 and 186 aspect 2 : ends of box at 165 and 175 aspect 3 : median marked at 170 or ft (a) provided \(165 \lt\) (a) \(\lt 175\) (B2 for box plot with two aspects correct) (B1 for one aspect or correct quartiles and median identified) SC : B2 for all 5 values (153, 165, ‘170’, 175, 186) plotted
Mark scheme (c)
Answer
Mark
Notes
Two correct comparisons
2
B1 ft from (b) for a correct comparison of range or inter-quartile range eg. the range / iqr is smaller for group B than group A B1 ft from (b) for a correct comparison of median or upper quartile or lower quartile or minimum or maximum eg. the median in group A is greater than the median in group B
The number of days each battery lasted was recorded.
(a) The frequency table represents the results for type P.
Number of days, \(d\)
Frequency
\(0 \leqslant d \lt 40\)
2
\(40 \leqslant d \lt 80\)
9
\(80 \leqslant d \lt 120\)
26
\(120 \leqslant d \lt 160\)
45
\(160 \leqslant d \lt 200\)
18
On the grid, draw a cumulative frequency diagram to represent the data. [3 marks]
(b) For type Q,
the median was 126 days the interquartile range was 57 days.
Compare the number of days that types P and Q lasted.
Make one statement about the average and one statement about the spread.
Use statistical measures to support your statements. [4 marks]
Mark scheme (a)
Answer
Mark
Comments
2 11 37 82 100
B1
may be implied by plots
Correct cf diagram with points joined with a smooth curve or lines
B2ft
\(\pm \dfrac{1}{2}\) small square ft their cf values which must be increasing for B2 or B1 B1ft cf diagram with all points plotted at their heights but not at correct horizontal positions or cf diagram with all points plotted at correct horizontal positions with at least 4 of their heights correct or all points plotted correctly at their heights but cf diagram not drawn or drawn poorly
Additional guidance
Ignore diagram to the left of their (40, 2)
For B2ft the diagram must end at (200, their 100) unless followed by a horizontal line
Histogram only
Max B1B0
Histogram and cf diagram
Mark the cf diagram
Mark scheme (b)
Answer
Mark
Comments
Correct median for their cumulative frequency diagram
B1ft
ft their diagram which must be increasing \(\pm \dfrac{1}{2}\) small square
Correct comparison of their type P median with 126
B1ft
answers must be in context eg P lasts longer or P is better
Correct interquartile range for their cumulative frequency diagram
B1ft
ft their diagram which must be increasing \(\pm \dfrac{1}{2}\) small square
Correct comparison of their type P interquartile range with 57
B1ft
answers must be in context eg P is more consistent or P is more reliable or Q is more varied
Additional guidance
2nd and 4th marks - ignore irrelevant or incorrect statements alongside a correct statement as long as not contradictory
eg (P median = 132) P lasts longer by 8 (error) minutes
2nd B1
P’s average is bigger
2nd B0
P’s median is greater
2nd B0
P’s spread is smaller
4th B0
P’s IQR is lower
4th B0
Q is more spread out
4th B0
2nd and 4th marks can be awarded even if their diagram is not increasing
2nd and 4th marks can be awarded even if the methods used for P’s median and IQR are incorrect
17 The table shows information about the heights of 60 athletes.
Height, \(h\) (cm)
Frequency
\(150 \lt h \leqslant 160\)
4
\(160 \lt h \leqslant 170\)
12
\(170 \lt h \leqslant 180\)
35
\(180 \lt h \leqslant 190\)
7
\(190 \lt h \leqslant 200\)
2
(a) Complete the cumulative frequency table. [1 mark]
Height, \(h\) (cm)
Cumulative frequency
\(h \leqslant 150\)
0
\(h \leqslant 160\)
4
\(h \leqslant 170\)
16
\(h \leqslant 180\)
\(h \leqslant 190\)
\(h \leqslant 200\)
(b) Circle the class interval that contains the lower quartile. [1 mark]
\(150 \lt h \leqslant 160\)
\(160 \lt h \leqslant 170\)
\(170 \lt h \leqslant 180\)
\(180 \lt h \leqslant 190\)
(c) Draw a cumulative frequency diagram to represent the data. [2 marks]
(d) Estimate the number of the athletes whose height is more than 176 cm [2 marks]
Mark scheme (a)
Answer
Mark
Comments
51, 58 and 60
B1
Mark scheme (b)
Answer
Mark
Comments
\(160 \lt h \leqslant 170\)
B1
Mark scheme (c)
Answer
Mark
Comments
Points plotted with upper class boundaries and cf values condone (150, 0) omitted or incorrectly plotted for this mark only
B1ft
\(\pm\dfrac{1}{2}\) square ft their cumulative frequencies, which must be increasing ignore bars drawn if points clearly plotted
Smooth curve or polygon
B1ft
ft their 5 or 6 points (point with cf 0 may be omitted) must be increasing and not a single straight line
Additional guidance
For the second mark, the points must be evenly spaced accept an omission of the point with cf 0, but do not accept an incorrect starting point for the pattern of their points accept a horizontal line drawn from their final point, but do not accept a continuation of the curve or polygon
Points plotted at lower class boundaries or midpoints, but with correct smooth curve or polygon for their points
B0B1
Bars drawn with correct curve
B1B1
Bars drawn without curve but with correct points clearly plotted
B1B0
Bars drawn without correct curve or correct points plotted
B0B0
Mark scheme (d)
Answer
Mark
Comments
Alternative method 1
Vertical line drawn from 176 to curve or polygon
M1
implied by correct reading for their increasing curve or polygon or mark at correct place on their increasing curve or polygon or on the vertical axis \(\pm\dfrac{1}{2}\) square
Correct value for \(60 -\) their reading or correct value for their \(60 -\) their reading
A1ft
ft their increasing curve or polygon answer must be an integer their 60 must be from an increasing curve or polygon
Alternative method 2
\(2 + 7 + \dfrac{4}{10} \times 35\) or \(2 + 7 + 14\) or \(4 + 12 + \dfrac{6}{10} \times 35\) or \(4 + 12 + 21\) or 37
M1
23
A1
Additional guidance
In alternative method 1 condone the curve or polygon drawn only for the required section (170 – 180) as long as the cumulative frequencies are increasing throughout
Answer 23 not from alternative method 2 must match their graph
15 Here is some information about the test marks of 120 students.
Mark, \(m\)
\(0 < m \leqslant 10\)
\(10 < m \leqslant 20\)
\(20 < m \leqslant 30\)
\(30 < m \leqslant 40\)
\(40 < m \leqslant 50\)
Frequency
20
28
40
20
12
(a) Complete the cumulative frequency table. [1 mark]
Mark, \(m\)
\(m \leqslant 10\)
\(m \leqslant 20\)
\(m \leqslant 30\)
\(m \leqslant 40\)
\(m \leqslant 50\)
Cumulative frequency
20
48
(b) Draw a cumulative frequency graph. [2 marks]
(c) Students who scored 15 marks or fewer take another test.
Use your graph to estimate how many students take another test. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
(20 48) 88 108 120
B1
Mark scheme (b)
Answer
Mark
Comments
All 5 points plotted using upper class bounds and their cf values
M1
\(\pm \dfrac{1}{2}\) small square must be increasing
Smooth curve or polygon for their cf values
A1ft
\(\pm \dfrac{1}{2}\) small square must be increasing
Additional guidance
If (a) is correct, points should be at (10, 20), (20, 48), (30, 88), (40, 108) and (50, 120)
For A1, the graph should start at (0, 0) or (1, 0) or (10, 20)
For A1, the graph should end at \(m = 50\) unless it followed by a horizontal line adjoining (50, 120)
Histogram only
M0A0
Histogram and graph
Mark curve
Mark scheme (c)
Answer
Mark
Comments
Line from 15 marks to their graph
M1
\(\pm \dfrac{1}{2}\) small square implied by mark at correct place on the graph or on the vertical axis (but not on the horizontal axis) or by correct reading from their graph
Correct reading from their graph for 15 marks
A1ft
\(\pm \dfrac{1}{2}\) small square
Additional guidance
Correct reading for their graph, with or without evidence of using graph
M1A1
No graph in (b)
M0A0
For M1 and A1ft the domain of their graph must be at least \(10 \leqslant m \leqslant 20\) and their graph must be increasing in the domain \(10 \leqslant m \leqslant 50\) or from \(m = 10\) if their graph does not extend to \(m = 50\)
19 Type A batteries and type B batteries were tested.
The cumulative frequency diagram shows information about the battery life of type A.
(a) Estimate the interquartile range for type A. [2 marks]
(b) Estimate the number of type A batteries that had a battery life of more than 1600 hours. [1 mark]
(c) The box plot shows information about the battery life of type B.
On average, which type had the greater battery life?
Tick a box.
type A
type B
Using data from both diagrams, state how you chose your answer. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
300
B2
B1 1100 or 1400 seen
Mark scheme (b)
Answer
Mark
Comments
4
B1
Additional guidance
Ignore incorrect ‘units’ eg 4 people
B1
Mark scheme (c)
Answer
Mark
Comments
Ticks type B and gives valid reason
B2
eg valid reasons (median for A is) 1260 and (median for B is) 1300 median for B is 40 more (than A)
B1 no or incorrect decision and (median for A is) 1260 and (median for B is) 1300 or no or incorrect decision and median for B is 40 more (than A) or ticks type B and (median for B is) 1300 and (median for A is) 1230 or 1280 or ticks type B and B has a larger median (than A) (if one median given it must be correct)
Additional guidance
If median values are not given in the wording, look for values on the graph and box plot
Ticks type B but gives no valid reason
B0
Allow use of average or middle for median, or a correct description eg ‘top 50%’. Do not accept ‘mean’ or ‘mode’ or other statistical measures for median
Ignore comments about measures other than the median
22 Here is some information about the miles per gallon of 60 cars.
Miles per gallon, \(x\)
Frequency
\(40 \lt x \leqslant 50\)
6
\(50 \lt x \leqslant 60\)
16
\(60 \lt x \leqslant 70\)
28
\(70 \lt x \leqslant 80\)
10
(a) Draw a cumulative frequency graph. [3 marks]
(b) Use the graph to work out the interquartile range. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
(6) 22 50 60
B1
cumulative frequency values may be implied by points plotted (\(\pm\) 0.5 square)
Points plotted with upper class boundaries and cf values (\(\pm\) 0.5 square)
B1ft
ft their cumulative frequencies must be increasing
Smooth curve or polygon (\(\pm\) 0.5 square)
B1ft
ft their cumulative frequencies must be increasing and not a single straight line
Additional guidance
Graphs may start from their first plotted point or from (40, 0) If the points are plotted at mid-points, with a point at (45, 6), the graph may start at (35, 0) (\(\pm\) 0.5 square) If the points are plotted at the lower bounds, with a point at (40, 6), the graph may start at (0, 0)
Graph starting at (0, 0), but otherwise correct
B1B1B0
Graph plotted at mid-points or lower class boundaries, but otherwise correct
B1B0B1
Graph ascends or descends after \(x = 80\)
B0 for 3rd mark
Bars drawn as well as correct graph
B1B1B0
Bars drawn without correct graph
max B1
Mark scheme (b)
Answer
Mark
Comments
One correct mpg reading for their graph from cf of 15(.25) or 45(.75) or horizontal lines from 15(.25) and 45(.75) only to their graph or 15(.25) and 45(.75) indicated as the cf values for the quartiles
19 Here is some information about the marks of 60 students in a test.
Mark, \(m\)
Frequency
\(40 \lt m \leqslant 50\)
9
\(50 \lt m \leqslant 60\)
16
\(60 \lt m \leqslant 70\)
20
\(70 \lt m \leqslant 80\)
8
\(80 \lt m \leqslant 90\)
7
(a) On the grid, draw a cumulative frequency graph. [3 marks]
(b) Use your graph to estimate the lowest mark of the top 20% of students. [2 marks]
Mark scheme (a)
Answer
Mark
Comments
(9) 25 45 53 60
B1
cumulative frequencies May be implied by points plotted (\(\pm\) 0.5 square)
Points plotted with upper class boundaries and cf values (\(\pm\)0.5 square)
B1ft
ft their cumulative frequencies Must be increasing and not a single straight line
Smooth curve or polygon starting at correct point for their points and going through all their points (\(\pm\)0.5 square)
B1ft
ft their cumulative frequencies Must be increasing and not a single straight line
Additional guidance
Graphs may start from their first plotted point or from (40, 0) If they have plotted their points at mid-points, with point at (45, 9), their graph may start at (35, 0) Graph starting at (0, 0), but otherwise correct
B1B1B0
Curve plotted at mid-points or lower class boundaries, but otherwise correct
B1B0B1
Ignore the graph after \(m = 90\)
Bars drawn as well as correct graph
B1B1B0
Bars drawn without the correct graph
max B1
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1
\(60 - 0.2 \times 60\) or \(60 \times 0.8\) or 48
M1
oe implied by horizontal line from 48 on vertical axis
Correct reading from their increasing graph
A1ft
\(\pm\,\dfrac{1}{2}\) square
Alternative method 2
\(70 + \dfrac{3}{8} \times 10\)
M1
[73, 75]
A1
Additional guidance
The correct answer is likely to be [73, 75] from a correct graph
14 The cumulative frequency graph shows the distribution of the examination scores for 120 students.
(a) A student says that they had an examination score of 34 and so were in the top half of the 120 students.
Is the student correct? Give a reason for your decision. [2]
(b) The top 10% of the 120 students are awarded a commendation.
Find the minimum mark required for a commendation. [3]
Mark scheme (a)
Answer
Marks
Part marks and guidance
any evidence of reading (median) from cf = 60 or reading cf for score 34
M1
For M1 we need to see a line or mark on the graph or a figure allowing an error of ± ½ small square so for 60 allow reading from 59 – 61 giving 35 – 37 for 34 allow reading from 33 – 35 gives a reading of 51 – 57 figures may be in working, see appendix for some A1 statements
No and a correct statement with [34 and] 35 – 37 or with 51 – 57 and 60
A1
Appendix: exemplar responses for Q14(a)
Response
Mark
35 – 37 is half way and 34 is not high enough
1
34 is less than 35 – 37
1
34 is at [c.f.] 51 – 57 which is less than 60
1
34 is at [c.f.] 51 – 57 which is 42.5% - 47.5% [percentile] and is less than 50% [percentile]
1
It is 1 – 3 marks short of halfway …. (implies 35 – 37)
1
51 – 57 is under half of 120
0
Mark scheme (b)
Answer
Marks
Part marks and guidance
58
3
M2 for \(\frac{100 - 10}{100} \times 120\) oe or 108 or M1 for \(\frac{10}{100} \times 120\) oe or 12
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
15 The cumulative frequency graph shows the distribution of the ages of the members of a tennis club.
(a) The table summarises the ages of the members of a cycling club.
Age (a years)
\(0 \lt a \leqslant 20\)
\(20 \lt a \leqslant 30\)
\(30 \lt a \leqslant 40\)
\(40 \lt a \leqslant 50\)
\(50 \lt a \leqslant 70\)
\(70 \lt a \leqslant 80\)
Frequency
8
14
8
12
17
5
On the graph above, draw the cumulative frequency graph of the ages of the members of the cycling club. [5]
(b) Find out which club has younger members on average. Give evidence to support your decision. [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
Correct cumulative frequency curve
5
B2 for 8 22 30 42 59 64 can be implied from plots or M1 for cf calculated with one arithmetic error M2 for accurately plotting these frequencies against the upper group values or M1 for accurately plotting these frequencies against a group value condoning one error B1 for an accurate curve FTtheir frequencies and dep on increasing frequencies
tolerance for plotting and drawing the curve is ± \(\frac{1}{2}\) small square and ignore curve before (20, 8), if they draw bars award M1 if the bars are the same widths as the groups.
Mark scheme (b)
Answer
Marks
Part marks and guidance
cycling and medians [cycling] 40 to 43 [tennis] 51
2
FTtheir curve for medians providing it is increasing frequencies B1 for one correct median or a statement with one error
See appendix condone a statement which is equivalent and condone tennis median of 50 to 52 and condone ‘average’ for ‘median’
Appendix: exemplar responses for Q15(b)
Response
Mark
Cycling + curve is higher for smaller values
2
Cycling + they have more under 40
2
Cycling + have a higher cf at 30
2
Cycling + earlier end of graph is higher than tennis
2
Cycling + the curve for younger members is steeper
2
Tennis + tennis has less members under the age of 40
1
Tennis + tennis curve is below cycling curve for lower ages
1
Tennis + tennis curve is less steep for lower ages
15 80 cyclists take part in a race. A summary of their times is shown in the table.
Time (\(t\) minutes)
Frequency
\(20 < t \leqslant 25\)
5
\(25 < t \leqslant 30\)
15
\(30 < t \leqslant 35\)
24
\(35 < t \leqslant 40\)
25
\(40 < t \leqslant 45\)
7
\(45 < t \leqslant 50\)
4
(a) Complete the cumulative frequency table.
Time (\(t\) minutes)
Cumulative frequency
\(t \leqslant 25\)
5
\(t \leqslant 30\)
\(t \leqslant 35\)
\(t \leqslant 40\)
\(t \leqslant 45\)
\(t \leqslant 50\)
[2]
(b) Draw the cumulative frequency graph to show the information.
[3]
(c) Reece makes two comments about the times taken to complete the race. For each comment, decide if Reece is right or wrong and give a reason for your answer.
(i) \(\frac{3}{4}\) of the 80 cyclists took more than 30 minutes to complete the race. [2]
(ii) The longest time that any of the 80 cyclists took to complete the race must have been 50 minutes. [1]
Mark scheme (a)
Answer
Marks
Part marks and guidance
20, 44, 69, 76, 80
2
M1 for cf calculated with one arithmetic error
Allow 80 or their 80 FT
Mark scheme (b)
Answer
Marks
Part marks and guidance
Plots at right-hand end of intervals
1
Plots of frequencies scores 0 across the whole of (b)
Plots at correct heights
1
Tol ½ square FT if M1 in (a)
0 for bars at correct heights since must miss off one end
Join with smooth curve or straight line segments
1
Tol ½ square FT their ascending plots only Condone curve or absence of curve below \(t = 25\)
if cf graph as well as bars, ignore bars
Mark scheme (c)
Answer
Marks
Part marks and guidance
(i)
Sight of 20 or 60
1
For 2 marks must say right/correct/true not yes
Must be separate from the fraction comment Do not accept more than/less than 60/20
See Appendix
\(\frac{20}{80} = \frac{1}{4}\) or \(\frac{1}{4}\) of 80 = 20 \(\frac{60}{80} = \frac{3}{4}\) or \(\frac{3}{4}\) of 80 = 60
1
Accept equivalent in words
(ii)
Wrong/incorrect/false and correct reason
1
e.g. cannot be sure as exact data not given; it could be anywhere between 45 and 50 etc.
Appendix: exemplar responses for Q15(c)(i)
Reason
Judgement
Mark
Correct, ¾ of 80 is 60 and 20 cyclists completed within 30 minutes, leaving the remaining ¾ of cyclists taking over 30 minutes
1 1
True, ¾ of 80 = 60 and 60 cyclists took more than 30 minutes
1 1
Right, 20 cyclists completed the race in under 30 minutes and 20/80 = ¼.
1 1
Right, 60 cyclists took more than 30 minutes and ¾ of 80 = 60.
1 1
Wrong, 60 cyclists took more than 30 minutes and ¾ of 80 = 60.
Despite “wrong”, mark the comments but do not award full marks
1 0
Yes, 60 cyclists took more than 30 minutes and ¾ of 80 = 60.
Do not award 2 marks with “yes”
1 0
Correct, if you add the amount of times after 30 you get 60 and there’s 80 cyclists.
1 0
Right, 60 cyclists took more than 30 minutes.
1 0
Wrong, 60 cyclists took more than 30 minutes.
Despite “wrong”, mark the comment
1 0
True, the lower quartile is at 30mins
Implies 20
1 0
¾ of 80 = 60
0 1
Appendix: exemplar responses for Q15(c)(ii)
Reason
Judgement
Mark
Wrong, all we know is that it is between 45 and 50
Condone 46 and 50
1
Wrong, there is a range of values for the time so not definitely 50
1
Wrong, as the time is given as a range
1
Wrong, we are not given the accurate values
1
Wrong, it might be but we need to know the exact values to find out
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 …..
(a) The cumulative frequency graph shows the distribution of the heights of members of a rowing club.
(i) Find the median. [1]
(ii) Find the interquartile range. [2]
(iii) Calculate the percentage of the members who are at least 180 cm tall. [3]
(b) The histogram summarises the heights of the 153 members of a swimming club.
Which club has the greater median height? You must show all your working. [5]
Mark scheme (a)
Answer
Marks
Part marks and guidance
(i) 172
1
(ii) 16 to 17
2
B1 for 160 or 176 to 177 (may be written or indicated on graph, not just a line through it)
(iii) 16.6 to 16.7 or 17
3
B2 for [0].83[3…] or 83[.3…]% or [0].166… or [0].167 or [0].17 OR B1 for 100 (from graph) or 20 M1 for \(\frac{\textit{their }100}{120}\) [× 100] or \(\frac{\textit{their }20}{120}\) [× 100]
Mark scheme (b)
Answer
Marks
Part marks and guidance
76.5 or 77 and 102 or both 28 (or 14+14) and 74
Swimming club has a median in group 160 to 170 oe [Rowing club has median their 172] So rowing club [has higher median] oe FTtheir(a)(i) for conclusion
5
B1 for 76.5 or 77 M2 for 20 × 1.4 and 10 × 7.4 soi by 102 or both 28 (or 14+14) and 74 or M1 for 20 × 1.4 or 10 × 7.4 soi by 28 (or 14+14) or 74 Accept any correct alternative methods (e.g. 5 squares = 1 person) B1 for [swimming club has a median in group] 160 to 170 oe e.g. “\(\leqslant 170\)” (if they use a proportional calculation answer 166 to 167) A1dep on previous 4 marks for “rowing club [has higher median]” oe FTtheir(a)(i) for conclusion
15 Iqrah carries out a survey of 200 families in the north of England on their weekly spending on food. The cumulative frequency diagram summarises the results.
(a) Find
(i) the median, [1]
(ii) the interquartile range. [2]
(b) Iqrah says
15% of these families spent over £120.
Is her statement correct? State the evidence you have used in making your decision. [2]
(c) In a survey of 200 families in the south of England, the median weekly amount spent on food was £84 and the interquartile range was £28.
Make two comparisons between the weekly amounts spent on food in the north of England and the south of England. State the evidence you have used in making your comparisons.
1 [2] 2 [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
(i) 90
1
(ii) 22
2
M1 for [UQ = ]100 or [LQ = ] 77 to 79
Accept 21 to 23
Mark scheme (b)
Answer
Marks
Part marks and guidance
No with 18 to 20 and 30 OR No with 8% to 10% [and 15%] OR No with [£] 110 to112 [which is less than 120] OR No with 170 and 180 to 184
2
M1 for 18 to 20 or 8% to 10% or 110 to 112 or for 30 or 170 or 180 to 184
Could be written on graph for M1
Mark scheme (c)
Answer
Marks
Part marks and guidance
Families in the south spent less on average as their median was lower oe
2
Strict FT their median in (a)(i) M1 for Families in the South spent less oe nfww
Allow either way around but do not allow M1 if wrong reason given e.g. in first reason mentions IQR for spending less Ignore ref to figures
Families in the south were more spread in their spending as their IQR was larger oe
2
Strict FT their IQR in (a)(ii) M1 for Spending varies more in the South oe nfww
For M1 allow spread oe associated with IQR without comparison
