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Either paper 3 marks
The frequency polygon shows the times of some shoppers.
(a)Write down the number of shoppers in the class \(20 < t \le 40\).
(b)Find the total number of shoppers.
Show answer
(a) \(9\)
(b) \(42\)Each point is plotted at the middle of its class. The middle of \(20\) to \(40\) is \(30\).
The point above \(30\) is at a height of 9, so there are 9 shoppers in that class.
Read the height of each of the five points: 7, 9, 14, 8 and 4.
\(7 + 9 + 14 + 8 + 4 = 42\)
- (b) M1reads the five heights \(7\), \(9\), \(14\), \(8\), \(4\) and adds them
- (b) A1total \(42\)
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Either paper 3 marks
The table gives information about the lengths of some fish.
Length (cm) Frequency \(0 < L \le 10\) 2 \(10 < L \le 20\) 10 \(20 < L \le 30\) 19 \(30 < L \le 40\) 9 \(40 < L \le 50\) 2 A frequency polygon has been started. The first two points are plotted.
(a)Write down the coordinates of the next point to plot.
(b)Work out the total number of fish.
Show answer
(a) \((25, 19)\)
(b) \(42\)A frequency polygon is plotted at the middle of each class.
The next class is \(20 < L \le 30\), and its middle is \(\frac{ 20 + 30 }{2} = 25\).
Its frequency is 19, so the point is \((25, 19)\).
\(2 + 10 + 19 + 9 + 2 = 42\)
- (a) M1x-coordinate at the middle of the class, \(25\), or y-coordinate 19
- (a) A1coordinates \((25, 19)\)
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Either paper 4 marks
33 Team A students and 33 Team B students each did a quiz.
The frequency polygons show how long they took.
(a)How many Team B students took between \(40\) and \(60\) minutes?
(b)How many more Team A students than Team B students took between \(20\) and \(40\) minutes?
(c)Which group generally took longer?
Show answer
(a) \(14\)
(b) \(9\)
(c) Team BThe middle of the class \(40\) to \(60\) is \(50\).
The Team B polygon is at a height of 14 above \(50\).
Both polygons are read at \(30\), the middle of that class.
Team A: 14 students. Team B: 5 students.
\(14 - 5 = 9\)
Both groups have 33 students, so the polygons can be compared directly.
The Team A polygon peaks at \(30\) minutes and the Team B polygon peaks later, at \(50\) minutes, so the Team B students generally took longer.
- (b) M1reads 14 and 5 at \(30\) and subtracts
- (b) A1difference \(9\)
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Either paper 3 marks
The frequency polygon shows the times of 37 visitors.
(a)Write down the modal class.
(b)Work out the number of visitors with a time in the queue of more than \(30\) minutes.
Show answer
(a) \(20 < t \le 30\)
(b) \(7\)The modal class is the class with the highest frequency, so look for the highest point.
It is plotted at \(25\), which is the middle of the class \(20 < t \le 30\).
\(30\) is a class boundary, so add the frequencies of the two classes above it.
Reading the last two points gives 4 and 3.
\(4 + 3 = 7\)
- (b) M1adds the frequencies 4 and 3 of the two classes above \(30\)
- (b) A1total \(7\)
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Either paper 2 marks
The table shows information about the time taken by some students to finish a puzzle.
Time (\(t\) minutes) Frequency \(0 < t \le 20\) 10 \(20 < t \le 40\) 14 \(40 < t \le 60\) 12 \(60 < t \le 80\) 18 \(80 < t \le 100\) 2 On the grid, draw a frequency polygon for this information. Plot the five points and join them with straight lines. Do not extend the polygon to the horizontal axis.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
Points at \((10, 10)\), \((30, 14)\), \((50, 12)\), \((70, 18)\) and \((90, 2)\), joined by straight linesA frequency polygon is plotted at the middle of each class, not at a class boundary.
The middle of \(0 < t \le 20\) is \(\frac{ 0 + 20 }{2} = 10\), so the first point is \((10, 10)\).
The middle of \(20 < t \le 40\) is \(\frac{ 20 + 40 }{2} = 30\), so the second point is \((30, 14)\).
The middle of \(40 < t \le 60\) is \(\frac{ 40 + 60 }{2} = 50\), so the third point is \((50, 12)\).
The middle of \(60 < t \le 80\) is \(\frac{ 60 + 80 }{2} = 70\), so the fourth point is \((70, 18)\).
The middle of \(80 < t \le 100\) is \(\frac{ 80 + 100 }{2} = 90\), so the fifth point is \((90, 2)\).
Join the five points with straight lines, and stop at the first and last points, as the question says.
- B1at least four of the five points plotted at the middle of their class, at the right frequency
- B1all five points plotted correctly and joined with straight lines
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Either paper 2 marks
The table shows information about the lengths of some video calls.
Length of call (\(L\) minutes) Frequency \(10 < L \le 30\) 1 \(30 < L \le 50\) 1 \(50 < L \le 70\) 1 \(70 < L \le 90\) 3 \(90 < L \le 110\) 4 \(110 < L \le 130\) 4 On the grid, draw a frequency polygon for the information in the table. Do not extend the polygon to the horizontal axis.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
Points at \((20, 1)\), \((40, 1)\), \((60, 1)\), \((80, 3)\), \((100, 4)\) and \((120, 4)\), joined by straight linesA frequency polygon is plotted at the middle of each class. The middle is the lower boundary plus the upper boundary, divided by 2.
\(\frac{ 10 + 30 }{2} = 20\), so the first point is \((20, 1)\).
\(\frac{ 30 + 50 }{2} = 40\), so the second point is \((40, 1)\).
\(\frac{ 50 + 70 }{2} = 60\), so the third point is \((60, 1)\).
\(\frac{ 70 + 90 }{2} = 80\), so the fourth point is \((80, 3)\).
\(\frac{ 90 + 110 }{2} = 100\), so the fifth point is \((100, 4)\).
\(\frac{ 110 + 130 }{2} = 120\), so the sixth point is \((120, 4)\).
Join the six points with straight lines, and stop at the first and last points, as the question says.
- B1at least five of the six points plotted at the middle of their class, at the right frequency
- B1all six points plotted correctly and joined with straight lines
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Either paper 4 marks
The frequency polygon shows the times of \(46\) students.
Each point is plotted at the middle of its class.
(a)Write down the modal class.
(b)Find an estimate for the mean time of the students.
Show answer
(a) \(40 < t \le 60\)
(b) \(50\ \text{minutes}\)The modal class is the one with the highest frequency, which is the highest point on the polygon: \(17\) students.
That point is plotted at \(50\), the middle of the class \(40 < t \le 60\).
The polygon does not say what any individual time was, only which class each student fell in — so the best that can be done is an estimate, using the middle of each class as though every student in it sat there.
The middles are already the \(x\)-coordinates of the plotted points.
\(10 \times 4 = 40\)
\(30 \times 10 = 300\)
\(50 \times 17 = 850\)
\(70 \times 12 = 840\)
\(90 \times 3 = 270\)
Adding those gives \(2300\), shared between \(46\) students.
\(2300 \div 46 = 50\) minutes
- (b) M1class middles times frequencies, e.g. \(50 \times 17 = 850\)
- (b) M1adding the products \(2300\) and dividing by \(46\)
- (b) A1estimate \(50\) minutes
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Either paper 3 marks
The table shows information about the lengths of 82 fish.
Length (\(L\) cm) Frequency \(7 < L \le 17\) 20 \(17 < L \le 27\) 17 \(27 < L \le 37\) 6 \(37 < L \le 47\) 16 \(47 < L \le 57\) 15 \(57 < L \le 67\) 8 (a)Find the class interval that contains the median.
(b)On the grid, draw a frequency polygon for the information in the table. Plot the six points and join them with straight lines. Do not extend the polygon to the horizontal axis.
You cannot draw on this page. Draw (b) on paper, then show the answer and check your drawing against the model.
Show answer
(a) \(27 < L \le 37\)
(b) Points at \((12, 20)\), \((22, 17)\), \((32, 6)\), \((42, 16)\), \((52, 15)\), \((62, 8)\), joined by straight lines(b)There are \(82\) fish, so the median is halfway through the ordered values, at position \(\frac{ 82 }{2} = 41\).
Build a running total of the frequencies:
\(20, \; 37, \; 43, \; 59, \; 74, \; 82\)
The running total is \(37\) before this class and \(37 + 6 = 43\) after it, so it passes \(41\) inside this class.
The median is in the class \(27 < L \le 37\).
A frequency polygon is plotted at the middle of each class, not at a class boundary.
The middle of \(7 < L \le 17\) is \(\frac{ 7 + 17 }{2} = 12\), so the first point is \((12, 20)\).
The middle of \(17 < L \le 27\) is \(\frac{ 17 + 27 }{2} = 22\), so the second point is \((22, 17)\).
The middle of \(27 < L \le 37\) is \(\frac{ 27 + 37 }{2} = 32\), so the third point is \((32, 6)\).
The middle of \(37 < L \le 47\) is \(\frac{ 37 + 47 }{2} = 42\), so the fourth point is \((42, 16)\).
The middle of \(47 < L \le 57\) is \(\frac{ 47 + 57 }{2} = 52\), so the fifth point is \((52, 15)\).
The middle of \(57 < L \le 67\) is \(\frac{ 57 + 67 }{2} = 62\), so the sixth point is \((62, 8)\).
Join the points with straight lines, from left to right.
- (a) B1the class \(27 < L \le 37\)
- (b) B1at least five points plotted at the middle of their classes, at the right height
- (b) B1all six points correct and joined with straight lines
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Either paper 3 marks
The frequency polygon shows the times of some shoppers.
(a)State the number of shoppers in the class \(80 < t \le 100\).
(b)Work out the total number of shoppers.
Show answer
(a) \(3\)
(b) \(48\)Each point is plotted at the middle of its class. The middle of \(80\) to \(100\) is \(90\).
The point above \(90\) is at a height of 3, so there are 3 shoppers in that class.
Read the height of each of the five points: 8, 14, 18, 5 and 3.
\(8 + 14 + 18 + 5 + 3 = 48\)
- (b) M1reads the five heights \(8\), \(14\), \(18\), \(5\), \(3\) and adds them
- (b) A1total \(48\)
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Either paper 3 marks
The table shows information about the times of some students.
Time (minutes) Frequency \(0 < t \le 10\) 7 \(10 < t \le 20\) 13 \(20 < t \le 30\) 15 \(30 < t \le 40\) 3 \(40 < t \le 50\) 1 A frequency polygon has been started. The first two points are plotted.
(a)Write down the coordinates of the next point to plot.
(b)Find the total number of students.
Show answer
(a) \((25, 15)\)
(b) \(39\)A frequency polygon is plotted at the middle of each class.
The next class is \(20 < t \le 30\), and its middle is \(\frac{ 20 + 30 }{2} = 25\).
Its frequency is 15, so the point is \((25, 15)\).
\(7 + 13 + 15 + 3 + 1 = 39\)
- (a) M1x-coordinate at the middle of the class, \(25\), or y-coordinate 15
- (a) A1coordinates \((25, 15)\)