Frequency Polygons (Part 2) Worksheet with Answers

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2.13B Frequency Polygons (Part 2)

EDEXCEL 1MA1 · Calculator allowed · about 80 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

1

A survey recorded how long, in minutes, 20 customers spent queueing at a supermarket checkout. The raw data is shown below.

3, 8, 12, 15, 22, 6, 17, 25, 11, 19, 9, 14, 28, 21, 5, 16, 23, 10, 18, 27

The grouped frequency table below is being completed from this data.

Queueing time (minutes)	Frequency
0-9	5
10-19	?
20-29	?

(a)Use the raw data to complete the frequency column for the classes 10-19 and 20-29.(2)

(b)Write down the modal class.(1)

(Total for Question 1 is 3 marks)

2

A farm shop records the mass, m kg, of bags of onions sold. The class intervals used are 0.5 ≤ m < 1.0, 1.0 ≤ m < 1.5, 1.5 ≤ m < 2.0 and 2.0 ≤ m < 2.5.

(a)Write down the midpoint of the class 0.5 ≤ m < 1.0.(1)

(b)Write down the midpoint of the class 1.5 ≤ m < 2.0.(1)

(Total for Question 2 is 2 marks)

3

Which of these statements about drawing a frequency polygon from a grouped frequency table is FALSE?

(a)Which of these statements about drawing a frequency polygon from a grouped frequency table is FALSE?(1)

A) Each point is plotted at the midpoint of its class interval.
B) Consecutive points are joined using straight lines.
C) The frequency polygon should be joined down to zero on the horizontal axis at both the first and last points.
D) Two frequency polygons can be plotted on the same axes to compare two distributions.

(Total for Question 3 is 1 mark)

4

A frequency polygon is being compared with a table of raw data for the same survey.

(a)Give a reason why a frequency polygon might be preferred to a table of raw data when displaying grouped continuous data.(1)

(Total for Question 4 is 1 mark)

5

30 students were asked how many text messages they sent yesterday. The grouped frequency table is shown, but one frequency is missing.

Number of texts (n)	Frequency
0 ≤ n < 10	6
10 ≤ n < 20	11
20 ≤ n < 30	y
30 ≤ n < 40	5

Total = 30

(a)Work out the value of y.(2)

(Total for Question 5 is 2 marks)

6

The table shows the distance, d km, cycled by 40 members of a cycling club on a Sunday ride.

Distance (d km)	Frequency	Midpoint
10 ≤ d < 15	7	?
15 ≤ d < 20	13	?
20 ≤ d < 25	12	?
25 ≤ d < 30	8	?

(a)Complete the midpoint column.(2)

(b)Write down the class width used in this table.(1)

(Total for Question 6 is 3 marks)

7

Using the table and midpoints from Question 6, draw a frequency polygon on a grid with 'Distance cycled (km)' on the horizontal axis from 0 to 30 and 'Frequency' on the vertical axis from 0 to 14.

Figure (to be drawn): Grid: horizontal axis 'Distance cycled (km)' from 0 to 30, with gridlines suitable for plotting to the nearest 0.5 km; vertical axis 'Frequency' from 0 to 14, gridlines every 1. No points pre-plotted; the grid is for the student to plot and join the frequency polygon using the completed table from Question 6.

(a)Plot the points (midpoint, frequency) for each class and join them in order with straight lines.(2)

(b)Ahmed joins the first and last points of his frequency polygon down to zero on the horizontal axis. Explain why this is not correct.(1)

(c)Write down the modal class.(1)

(Total for Question 7 is 4 marks)

8

A frequency polygon for the waiting time, t minutes, of 40 customers at a post office is formed by plotting the midpoint of each class against its frequency. The polygon passes through the points (2, 7), (6, 13), (10, 10), (14, 6), (18, 4), where each class has a width of 4 minutes.

(a)Write down the frequency of the class with midpoint 6 minutes.(1)

(b)Work out the number of customers who waited 12 minutes or more.(2)

(c)Write down the modal class.(1)

(Total for Question 8 is 4 marks)

9

A theme park records the time, t minutes, that visitors spend queueing for a ride, using class intervals of equal width covering the range 0 to 60 minutes. The table currently uses a class width of 12 minutes.

(a)Work out how many extra class intervals would be needed if the class width were changed from 12 minutes to 6 minutes, using the same 0 to 60 minute range.(2)

(Total for Question 9 is 2 marks)

10

The table shows the daily screen time, m minutes, of 30 teenagers surveyed for a school project.

Screen time (m minutes)	Frequency
0 ≤ m < 20	4
20 ≤ m < 40	9
40 ≤ m < 60	10
60 ≤ m < 80	5
80 ≤ m < 100	2

(a)Calculate an estimate for the mean screen time.(3)

(Total for Question 10 is 3 marks)

11

The table shows the height, h cm, of 30 sunflowers grown in a school garden.

Height (h cm)	Frequency
0 ≤ h < 20	3
20 ≤ h < 40	9
40 ≤ h < 60	14
60 ≤ h < 80	4

Priya is drawing a frequency polygon for this table. Instead of using the midpoint of each class, she plots each point using the upper class boundary, giving the points (20, 3), (40, 9), (60, 14), (80, 4).

(a)Explain what is wrong with Priya's method.(1)

(b)Write down the four points Priya should have plotted.(2)

(Total for Question 11 is 3 marks)

12

For each statement about frequency polygons, write True or False.
(i) A frequency polygon can be drawn directly from a grouped frequency table without first finding the midpoint of each class.
(ii) Two frequency polygons can be plotted on the same axes to compare two sets of data, provided both tables use the same class widths.

(i)A frequency polygon can be drawn directly from a grouped frequency table without first finding the midpoint of each class.(1)

(ii)Two frequency polygons can be plotted on the same axes to compare two sets of data, provided both tables use the same class widths.(1)

(Total for Question 12 is 2 marks)

13

The frequency polygons for the 100 m swim times, s seconds, of 30 members at Riverside Swimming Club and 30 members at Oakwood Swimming Club are shown below, using classes of width 5 seconds.
Riverside SC passes through: (52.5, 4), (57.5, 9), (62.5, 11), (67.5, 4), (72.5, 2)
Oakwood SC passes through: (52.5, 8), (57.5, 12), (62.5, 6), (67.5, 3), (72.5, 1)

(a)Write down the modal class for Riverside SC.(1)

(b)Write down the modal class for Oakwood SC.(1)

(c)Erin says, 'Oakwood's swimmers are generally faster than Riverside's.' Use the frequency polygons to decide whether Erin is correct, supporting your answer with a calculation.(3)

(Total for Question 13 is 5 marks)

14

The frequency polygon for the time, r minutes, spent revising per day by 25 students passes through the points (5, 3), (15, 7), (25, 9), (35, 4), (45, 2), where each class has a width of 10 minutes.

(a)Work out the percentage of students who revised for at least 20 minutes per day.(3)

(Total for Question 14 is 3 marks)

15

The table shows the age, y years, of 30 visitors to a museum exhibit.

Age (y years)	Frequency
15 ≤ y < 25	6
25 ≤ y < 35	10
35 ≤ y < 45	8
45 ≤ y < 55	6

(a)Use the class boundaries to find an estimate for the range of the ages of the visitors.(2)

(Total for Question 15 is 2 marks)

16

A frequency polygon for the length, in minutes, of films watched by 50 members of a cinema club passes through the points (10, 5), (30, 2x), (50, 3x), (70, x) and (90, 3), where each class has a width of 20 minutes.

(a)Work out the value of x.(3)

(Total for Question 16 is 3 marks)

17

24 participants in a parkrun recorded the distance, in km, they had run in total that month.

1, 3, 5, 6, 7, 7, 4, 9, 10, 8, 11, 9, 10, 8, 9, 13, 14, 12, 15, 14, 13, 17, 19, 22

The raw data is being grouped into a frequency table using classes of width 4 km, starting at 0.

Distance (d km)	Frequency
0 ≤ d < 4	2
4 ≤ d < 8	5
8 ≤ d < 12	?
12 ≤ d < 16	?
16 ≤ d < 20	?
20 ≤ d < 24	1

Figure (to be drawn): Grid: horizontal axis 'Distance run (km)' from 0 to 24, gridlines suitable for plotting to the nearest km; vertical axis 'Frequency' from 0 to 9, gridlines every 1. No points pre-plotted; the grid is for the student to plot and join the frequency polygon using the completed table from part (a).

(a)Use the raw data to complete the frequency column for the classes 8 ≤ d < 12, 12 ≤ d < 16 and 16 ≤ d < 20.(2)

(b)Using midpoints, draw a frequency polygon for this data on the grid provided.(2)

(c)Write down the modal class.(1)

(d)Work out the number of participants who ran 12 km or more in total that month.(1)

(Total for Question 17 is 6 marks)

18

Two coach companies each recorded the journey time, t minutes, of 40 trips, using classes of width 10 minutes.
Company X passes through: (15, 2), (25, 10), (35, 18), (45, 8), (55, 2)
Company Y passes through: (15, 5), (25, 9), (35, 12), (45, 10), (55, 4)

(a)Compare the spread of journey times for the two companies. Justify your answer by comparing the number of trips in the fastest class (10 ≤ t < 20) and the slowest class (50 ≤ t < 60) for each company.(3)

(Total for Question 18 is 3 marks)

19

Using the data for Company X from Question 18, calculate an estimate for the mean journey time.

(a)Calculate an estimate for the mean journey time for Company X.(3)

(Total for Question 19 is 3 marks)

20

Using the data for Company Y from Question 18, Company Y advertises: 'The vast majority of our journeys take under 45 minutes.'

(a)Work out the percentage of Company Y's journeys that took at least 45 minutes.(2)

(b)Using your answer to part (a), comment on whether Company Y's claim is justified.(2)

(Total for Question 20 is 4 marks)

Mark scheme · 2.13B Frequency Polygons (Part 2)

Question 1

(a) B1 10-19 frequency = 9
(a) B1 20-29 frequency = 6
(a) Answer: 9 and 6
(b) B1 10-19 cao
(b) Answer: 10-19

Question 2

(a) B1 0.75 cao
(a) Answer: 0.75
(b) B1 1.75 cao
(b) Answer: 1.75

Question 3

(a) B1 C cao
(a) Answer: C) The frequency polygon should be joined down to zero on the horizontal axis at both the first and last points.

Question 4

(a) B1 any valid reason, e.g. it shows the overall shape/pattern of the data clearly, or it makes it easy to compare two or more distributions on the same axes oe
(a) Answer: e.g. it shows the overall shape of the distribution more clearly than a list of raw numbers.

Question 5

(a) M1 30 - (6+11+5) oe
(a) A1 y = 8 cao
(a) Answer: 8

Question 6

(a) M1 correct method shown for at least one midpoint, e.g. (10+15)/2
(a) A1 all four midpoints correct: 12.5, 17.5, 22.5, 27.5
(a) Answer: 12.5, 17.5, 22.5, 27.5
(b) B1 5 (km) cao
(b) Answer: 5 km

Question 7

(a) M1 all four points plotted correctly at (12.5,7), (17.5,13), (22.5,12), (27.5,8) ft from Q6(a)
(a) A1 points joined in the correct order with straight lines using a ruler
(a) Answer: Frequency polygon through (12.5,7), (17.5,13), (22.5,12), (27.5,8)
(b) B1 correct explanation, e.g. a frequency polygon only joins the plotted midpoint-frequency points with straight lines; joining down to the axis is not part of the convention and misrepresents data outside the given classes oe
(b) Answer: The polygon should only join the plotted points; it should not be extended down to the horizontal axis.
(c) B1 15 ≤ d < 20 cao
(c) Answer: 15 ≤ d < 20

Question 8

(a) B1 13 cao
(a) Answer: 13
(b) M1 6 + 4 oe (classes with midpoint 14 and 18)
(b) A1 = 10 cao
(b) Answer: 10
(c) B1 4 ≤ t < 8 cao
(c) Answer: 4 ≤ t < 8

Question 9

(a) M1 60/12=5 oe and 60/6=10 oe (both counts of class intervals found)
(a) A1 5 (extra class intervals) cao
(a) Answer: 5 extra class intervals

Question 10

(a) M1 correct midpoints identified (10, 30, 50, 70, 90) used to find at least 3 correct fx products
(a) M1 sum of fx = 1340
(a) A1 mean = 1340/30 = awrt 44.7 (minutes) cao
(a) Answer: awrt 44.7 minutes

Question 11

(a) B1 correct explanation, e.g. she should use the midpoint of each class, not the upper class boundary, because the midpoint represents the whole class oe
(a) Answer: She should have used the midpoint of each class, not the upper class boundary.
(b) M1 midpoint found using (lower + upper)/2 for at least one class
(b) A1 all four points correct: (10,3), (30,9), (50,14), (70,4)
(b) Answer: (10,3), (30,9), (50,14), (70,4)

Question 12

(i) B1 False cao
(i) Answer: False
(ii) B1 True cao
(ii) Answer: True

Question 13

(a) B1 60 ≤ s < 65 cao
(a) Answer: 60 ≤ s < 65
(b) B1 55 ≤ s < 60 cao
(b) Answer: 55 ≤ s < 60
(c) M1 correct classes under 60 seconds identified and frequencies summed for at least one club
(c) A1 Riverside: 4+9=13 out of 30 (43.3%); Oakwood: 8+12=20 out of 30 (66.7%) oe
(c) C1 valid conclusion consistent with the figures, e.g. Erin is correct because a much greater proportion of Oakwood's times are under 60 seconds oe
(c) Answer: Erin is correct; Oakwood's swim times are generally faster.

Question 14

(a) M1 identify the correct classes (midpoint 25, 35, 45) and sum the frequencies: 9+4+2=15
(a) M1 15/25 oe
(a) A1 60% cao
(a) Answer: 60%

Question 15

(a) M1 55 - 15 oe
(a) A1 = 40 (years) cao
(a) Answer: 40 years

Question 16

(a) M1 5 + 2x + 3x + x + 3 = 50 oe
(a) dM1 6x + 8 = 50 leading to 6x = 42
(a) A1 x = 7 cao
(a) Answer: x = 7

Question 17

(a) M1 correct count shown or implied for at least one of the three missing classes
(a) A1 all three frequencies correct: 8, 6, 2
(a) Answer: 8, 6, 2
(b) M1 all six points plotted correctly at (2,2), (6,5), (10,8), (14,6), (18,2), (22,1) ft from part (a)
(b) A1 points joined in the correct order with straight lines using a ruler
(b) Answer: Frequency polygon through (2,2), (6,5), (10,8), (14,6), (18,2), (22,1)
(c) B1 8 ≤ d < 12 cao
(c) Answer: 8 ≤ d < 12
(d) B1 9 ft from part (a): 6+2+1
(d) Answer: 9

Question 18

(a) C1 Company X: 2+2=4 trips combined in the fastest and slowest classes; Company Y: 5+4=9 trips oe
(a) C1 correct comparison: Company Y has more trips in the extreme classes than Company X
(a) C1 valid conclusion, e.g. Company Y's journey times are more spread out (more varied), while Company X's times are more concentrated around the middle classes oe
(a) Answer: Company Y's journey times are more spread out than Company X's.

Question 19

(a) M1 at least 3 correct fx products: 30, 250, 630, 360, 110
(a) M1 sum of fx = 1380
(a) A1 mean = 1380/40 = 34.5 (minutes) cao
(a) Answer: 34.5 minutes

Question 20

(a) M1 10+4=14 identified as the classes with midpoint 45 and 55
(a) A1 35% cao (14/40)
(a) Answer: 35%
(b) C1 100-35=65% of journeys take under 45 minutes oe
(b) C1 valid judgement consistent with 65%, e.g. accept that 65% is a majority but arguably not a 'vast' majority, so the claim is questionable oe; or a reasoned argument that 65% does justify the claim, provided it is supported by the figure
(b) Answer: 65% of journeys take under 45 minutes; whether this counts as a 'vast majority' is debatable, so the claim is only partly justified.

Frequency Polygons (Part 2) - Worksheets, Questions and Revision

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2.13B Frequency Polygons (Part 2)

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Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

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