MEP Pupil Text 9
9.4
But the length of the second interval is 10.5 − 5.5 = 5 , so the median is estimated by
8.5
× 5 = 3.86
11
from the start of this interval. Therefore the median is estimated as
5.5 + 3.86 = 9.36 .
(c)
The modal class is 1–5, as this class contains the most entries.
Exercises
1.
A door to door salesman keeps a record of the number of homes he visits each day.
Homes visited
Frequency
(a)
(b)
(c)
2.
20 – 29
30 – 39
40 – 49
3
8
24
60
21
The weights of a number of students were recorded in kg.
30 ≤ w < 35 35 ≤ w < 40
40 ≤ w < 45
45 ≤ w < 50
50 ≤ w < 55
11
15
7
4
(a)
Estimate the mean weight.
(b)
(c)
What is the modal class?
Frequency
10
Estimate the median.
A stopwatch was used to find the time that it took a group of children to run 100 m.
Time (seconds)
Frequency
(a)
(c)
(d)
4.
10 – 19
Estimate the mean number of homes visited.
Estimate the median.
What is the modal class?
Mean (kg)
3.
0–9
10 ≤ t < 15
15 ≤ t < 20
20 ≤ t < 25
25 ≤ t < 30
6
16
21
8
Is the median in the modal class?
(b) Estimate the mean.
Estimate the median.
Is the median greater or less than the mean?
The distances that children in a year group travelled to school is recorded.
Distance (km)
Frequency
(a)
(b)
(c)
0 ≤ d < 0.5
30
0.5 ≤ d < 1.0
1.0 ≤ d < 1.5 1.5 ≤ d < 2.0
22
Does the modal class contain the median?
Estimate the median and the mean.
Which is the largest, the median or the mean?
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19
8
MEP Pupil Text 9
5.
The ages of the children at a youth camp are summarised in the table below.
Age (years)
6–8
9 – 11
12 – 14
15 – 17
Frequency
8
22
29
5
Estimate the mean age of the children.
6.
The lengths of a number of leaves collected for a project are recorded.
Length (cm)
2–5
6 – 10
11 – 15
16 – 25
8
20
42
12
Frequency
Estimate (a) the mean
7.
(b) the median length of a leaf.
The table shows how many nights people spend at a campsite.
Number of nights
1–5
Frequency
8.
6 – 10
20
11 – 15
26
16 – 20
32
5
2
(a)
Estimate the mean.
(c)
What is the modal class?
(a)
A teacher notes the number of correct answers given by a class on
a multiple-choice test.
Correct answers
Frequency
(i)
(iii)
(b)
Estimate the median.
1 – 10
11 – 20
21 – 30
2
8
15
Estimate the mean.
What is the modal class?
(ii)
31 – 40
11
41 – 50
3
Estimate the median.
Another class took the same test. Their results are given below.
Correct answers
Frequency
(i)
(iii)
(c)
(b)
21 – 25
1 – 10
11 – 20
21 – 30
31 – 40
41 – 50
3
14
20
2
1
Estimate the mean.
What is the modal class?
(ii)
Estimate the median.
How do the results for the two classes compare?
Information
A quartile is one of 3 values (lower quartile, median and upper quartile) which divides
data into 4 equal groups.
A percentile is one of 99 values which divides data into 100 equal groups.
The lower quartile corresponds to the 25th percentile. The median corresponds to the
50th percentile. The upper quartile corresponds to the 75th percentile.
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MEP Pupil Text 9
9.4
9.
29 children are asked how much pocket money they were given last week.
Their replies are shown in this frequency table.
Pocket money
£
10.
Frequency
f
0 – £1.00
12
£1.01 – £2.00
9
£2.01 – £3.00
6
£3.01 – £4.00
2
(a)
Which is the modal class?
(b)
Calculate an estimate of the mean amount of pocket money received per child.
(NEAB)
The graph shows the number of hours a sample of people spent viewing television
one week during the summer.
40
30
Number of
people
20
10
0
(a)
10
20
30
40
50
Viewing time (hours)
60
70
Copy and complete the frequency table for this sample.
Viewing time
(h hours)
Number of
people
0 ≤ h < 10
13
10 ≤ h < 20
27
20 ≤ h < 30
33
30 ≤ h < 40
40 ≤ h < 50
50 ≤ h < 60
(b)
Another survey is carried out during the winter. State one difference you
would expect to see in the data.
(c)
Use the mid-points of the class intervals to calculate the mean viewing time
for these people. You may find it helpful to use the table below.
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MEP Pupil Text 9
Viewing time
(h hours)
Mid-point
Frequency
Mid-point ×
Frequency
0 ≤ h < 10
5
13
65
10 ≤ h < 20
15
27
405
20 ≤ h < 30
25
33
825
30 ≤ h < 40
35
40 ≤ h < 50
45
50 ≤ h < 60
55
(SEG)
11.
In an experiment, 50 people were asked to estimate the length of a rod to the
nearest centimetre. The results were recorded.
Length (cm)
20
21
22
23
24
25
26
27
28
29
Frequency
0
4
6
7
9
10
7
5
2
0
(a)
Find the value of the median.
(b)
Calculate the mean length.
(c)
In a second experiment another 50 people were asked to estimate the length
of the same rod. The most common estimate was 23 cm. The range of the
estimates was 13 cm.
Make two comparisons between the results of the two experiments.
(SEG)
12.
The following list shows the maximum daily temperature, in °F , throughout the
month of April.
(a)
56.1
49.4
63.7
56.7
55.3
53.5
52.4
57.6
59.8
52.1
45.8
55.1
42.6
61.0
61.9
60.2
57.1
48.9
63.2
68.4
55.5
65.2
47.3
59.1
53.6
52.3
46.9
51.3
56.7
64.3
Copy and complete the grouped frequency table below.
Temperature, T
Frequency
40 < T ≤ 50
50 < T ≤ 54
54 < T ≤ 58
58 < T ≤ 62
62 < T ≤ 70
(b)
Use the table of values in part (a) to calculate an estimate of the mean of this
distribution. You must show your working clearly.
(c)
Draw a histogram to represent your distribution in part (a).
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(MEG)
MEP Pupil Text 9
(c)
120
108
Here the mark which was attained
by the top 10% is required.
100
10% of 120 = 12
80
Cumulative
Frequency
so start at 108 on the cumulative
frequency scale.
60
This gives a mark of 79.
40
20
79
0
20
0
40
60
80
100
120
Test Score
(d)
103
To find the number of students who
scored more than 75, start at 75 on
the horizontal axis.
100
80
This gives a cumulative frequency Cumulative
Frequency 60
of 103.
So the number of students with a
score greater than 75 is
40
20
120 − 103 = 17 .
75
0
20
0
40
60
80
100
Test Score
As in Worked Example 1, a more accurate estimate for the median and inter-quartile
range is obtained if you draw a smooth curve through the data points.
Exercises
1.
Make a cumulative frequency table for each set of data given below. Then draw a
cumulative frequency graph and use it to find the median and inter-quartile range.
(a)
John weighed each apple in a large box. His results are given in this table.
Weight of
apple (g)
60 < w ≤ 80 80 < w ≤ 100 100 < w ≤ 120 120 < w ≤ 140 140 < w ≤ 160
Frequency
(b)
4
28
33
27
8
Pasi asked the students in his class how far they travelled to school each day.
His results are given below.
Distance (km)
Frequency
0 < d ≤1 1< d ≤ 2
5
2<d≤3
3<d≤4
5
6
12
167
4<d≤5 5<d≤6
5
3
MEP Pupil Text 9
9.5
(c)
A P.E. teacher recorded the distances children could reach in the long jump
event. His records are summarised in the table below.
Length of jump (m)
1< d ≤2
2<d≤3
3<d≤4
5
12
5
Frequency
2.
6
5
A farmer grows a type of wheat in two different fields. He takes a sample of
50 heads of corn from each field at random and weighs the grains he obtains.
Mass of grain (g)
0<m≤5
Frequency Field A
3
8
22
Frequency Field B
0
11
34
3.
5 < m ≤ 10 10 < m ≤ 15 15 < m ≤ 20
20 < m ≤ 25
25 < m ≤ 30
10
4
3
4
1
0
(a)
Draw cumulative frequency graphs for each field.
(b)
Find the median and inter-quartile range for each field.
(c)
Comment on your results.
A consumer group tests two types of batteries using a personal stereo.
Lifetime (hours)
4.
4<d≤5 5<d≤6
2<l≤3 3<l≤4
4<l≤5 5<l≤6
6<l≤7 7<l≤8
Frequency Type A
1
3
10
22
8
4
Frequency Type B
0
2
2
38
6
0
(a)
Use cumulative frequency graphs to find the median and inter-quartile range
for each type of battery.
(b)
Which type of battery would you recommend and why?
The table below shows how the height of girls of a certain age vary. The data was
gathered using a large-scale survey.
Height (cm) 50 < h ≤ 55 55 < h ≤ 60 60 < h ≤ 65 65 < h ≤ 70 70 < h ≤ 75 75 < h ≤ 80 80 < h ≤ 85
Frequency
100
300
2400
1300
700
150
50
A doctor wishes to be able to classify children as:
Category
Percentage of Population
Very Tall
Tall
Normal
Short
5%
15%
60%
15%
Very short
5%
Use a cumulative frequency graph to find the heights of children in each category.
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MEP Pupil Text 9
5.
The manager of a double glazing company employs 30 salesmen. Each year he
awards bonuses to his salesmen.
Bonus
Awarded to
£500
£250
Best 10% of salesmen
Middle 70% of salesmen
£ 50
Bottom 20% of salesmen
The sales made during 1995 and 1996 are shown in the table below.
Value of sales
(£1000)
0 < V ≤ 100 100 < V ≤ 200 200 < V ≤ 300 300 < V ≤ 400 400 < V ≤ 500
Frequency 1996
0
2
15
10
3
Frequency 1995
2
8
18
2
0
Use cumulative frequency graphs to find the values of sales needed to obtain each
bonus in the years 1995 and 1996.
6.
The histogram shows the cost of buying a particular toy in a number of different
shops.
7
6
5
Frequency 4
3
2
1
0
2.00
(a)
(b)
2.20
2.40
2.60
Price (£)
2.80
3.00
Draw a cumulative frequency graph and use it to answer the following
questions.
(i)
How many shops charged more than £2.65?
(ii)
What is the median price?
(iii)
How many shops charged less than £2.30?
(iv)
How many shops charged between £2.20 and £2.60?
(v)
How many shops charged between £2.00 and £2.50?
Comment on which of your answers are exact and which are estimates.
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MEP Pupil Text 9
9.5
7.
Laura and Joy played 40 games of golf together. The table below shows Laura's
scores.
Scores (x)
70 < x ≤ 80 80 < x ≤ 90 90 < x ≤ 100 100 < x ≤ 110 110 < x ≤ 120
Frequency
(a)
1
4
15
17
3
On a grid similar to the one below, draw a cumulative frequency diagram to
show Laura's scores.
q
y
40
30
Cumulative
Frequency
20
10
60
70
80
90
100
110
120
Score
(b)
Making your method clear, use your graph to find
(i)
(ii)
(c)
8.
Laura's median score,
the inter-quartile range of her scores.
Joy's median score was 103. The inter-quartile range of her scores was 6.
(i)
Who was the more consistent player?
Give a reason for your choice.
(ii)
The winner of a game of golf is the one with the lowest score.
Who won most of these 40 games? Give a reason for your choice.
(NEAB)
A sample of 80 electric light bulbs was taken. The lifetime of each light bulb was
recorded. The results are shown below.
Lifetime (hours)
800–
900–
Frequency
4
13
Cumulative Frequency
4
17
1000–
17
1100–
22
1200–
1300–
1400–
20
4
0
(a)
Copy and complete the table of values for the cumulative frequency.
(b)
Draw the cumulative frequency curve, using a grid as shown below.
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MEP Pupil Text 9
90
y
80
70
60
q
Cumulative
50
Frequency
40
30
20
10
0
800
900
1000
1100
1200
1300
1400
1500
Lifetime (hours)
Use your graph to estimate the number of light bulbs which lasted more than
1030 hours.
(d)
Use your graph to estimate the inter-quartile range of the lifetimes of the
light bulbs.
(e)
A second sample of 80 light bulbs has the same median lifetime as the first
sample. Its inter-quartile range is 90 hours. What does this tell you about
the difference between the two samples?
(SEG)
The numbers of journeys made by a group of people using public transport in one
month are summarised in the table.
Number of journeys
Number of people
(a)
0–10
11–20
21–30
31–40
41–50
51–60
61–70
4
7
8
6
3
4
0
≤ 60
≤ 70
Copy and complete the cumulative frequency table below.
Number of journeys
≤10
≤ 20
≤ 30
≤ 40
≤ 50
Cumulative frequency
(b)
(i)
Draw the cumulative frequency graph, using a grid as below.
y
40
30
q
9.
(c)
Cumulative
Frequency
20
10
0
10
20
30
40
Number of journeys
171
50
60
70
MEP Pupil Text 9
9.5
(c)
(ii)
Use your graph to estimate the median number of journeys.
(iii)
Use your graph to estimate the number of people who made more than
44 journeys in the month.
The numbers of journeys made using public transport in one month, by
another group of people, are shown in the graph.
40
30
Cumulative
Frequency
20
10
0
10
20
30
40
Number of journeys
50
60
70
Make one comparison between the numbers of journeys made by these
two groups.
(SEG)
The cumulative frequency graph below gives information on the house prices in
1992. The cumulative frequency is given as a percentage of all houses in England.
100
y( )
80
60
q
10.
Cumulative
Frequency
40
20
0
100000
200000 240000
i (£)
House prices in 1992
This grouped frequency table gives the percentage distribution of house prices (p)
in England in 1993.
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MEP Pupil Text 9
House prices (p)
in pounds 1993
(a)
Percentage of houses
in this class interval
0 ≤ p < 40 000
26
40 000 ≤ p < 52 000
19
52 000 ≤ p < 68 000
22
68 000 ≤ p < 88 000
15
88 000 ≤ p < 120 000
9
120 000 ≤ p < 160 000
5
160 000 ≤ p < 220 000
4
Use the data above to complete the cumulative frequency table below.
House prices (p)
in pounds 1993
Cumulative
Frequency (%)
0 ≤ p < 40 000
0 ≤ p < 52 000
0 ≤ p < 68 000
0 ≤ p < 88 000
0 ≤ p < 120 000
0 ≤ p < 160 000
0 ≤ p < 220 000
(b)
Trace or photocopy the grid for 1992, and on it construct a cumulative
frequency graph for your table for 1993.
(c)
In 1992 the price of a house was £100 000. Use both cumulative frequency
graphs to estimate the price of this house in 1993. Make your method clear.
(LON)
11.
The lengths of a number of nails were measured to the nearest 0.01 cm, and the
following frequency distribution was obtained.
Length of nail
(x cm)
Number of nails
0.98 ≤ x < 1.00
2
1.00 ≤ x < 1.02
4
1.02 ≤ x < 1.04
10
1.04 ≤ x < 1.06
24
1.06 ≤ x < 1.08
32
1.08 ≤ x < 1.10
17
1.10 ≤ x < 1.12
7
1.12 ≤ x < 1.14
4
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Cumulative Frequency
MEP Pupil Text 9
9.5
(a)
Complete the cumulative frequency column.
(b)
Draw a cumulative frequency diagram on a grid similar to the one below.
100
80
60
Cumulative
Frequency
40
20
0
0.98
1.00
1.02
1.04
1.06
1.08
1.10
1.12
1.14
Length of nail (cm)
Use your graph to estimate
(i)
12.
the median length of the nails
(ii)
the inter-quartile range.
(NEAB)
A wedding was attended by 120 guests. The distance, d miles, that each guest
travelled was recorded in the frequency table below.
Distance
(d miles)
Number of
guests
0 < d ≤ 10 10 < d ≤ 20
26
38
20 < d ≤ 30
30 < d ≤ 50
20
20
50 < d ≤ 100
12
100 < d ≤ 140
4
(a)
Using the mid-interval values, calculate an estimate of the mean distance
travelled.
(b)
(i)
Distance
(d miles)
Copy and complete the cumulative frequency table below.
d ≤ 10
d ≤ 20
d ≤ 30
Number of
guests
(ii)
d ≤ 50
d ≤ 100
d ≤ 140
120
On a grid similar to that shown below, draw a cumulative frequency
curve to represent the information in the table.
Just for Fun
Comment on this statement. "Students who smoke perform worse academically than nonsmokers, so smoking will make you stupid."
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MEP Pupil Text 9
120
100
80
Cumulative
Frequency
60
40
20
0
20
40
60
80
100
120
140
Distance travelled (miles)
(c)
(i)
Use the cumulative frequency curve to estimate the median distance
travelled by the guests.
(ii)
Give a reason for the large difference between the mean distance and
the median distance.
(MEG)
9.6 Standard Deviation
The two frequency polygons drawn on the graph below show samples which have the
same mean, but the data in one are much more spread out than in the other.
25
20
Frequency
15
10
5
0
1
2
3
4
5
6
7
8
9
10
11
12
Length
The range (highest value – lowest value) gives a simple measure of how much the data
are spread out.
175