` 14 42 115 217 294 329 350 Cumulative Frequency How to draw a cumulative frequency graph.
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Transcript of ` 14 42 115 217 294 329 350 Cumulative Frequency How to draw a cumulative frequency graph.
![Page 1: ` 14 42 115 217 294 329 350 Cumulative Frequency How to draw a cumulative frequency graph.](https://reader035.fdocuments.net/reader035/viewer/2022062312/551ae9e9550346f70d8b4a75/html5/thumbnails/1.jpg)
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14
42
115217
294
329350
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Cumulative Frequency
How to draw a cumulative
frequency graph
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Data x Frequency10 ≤ x < 20
2
20 ≤ x < 30
4
30 ≤ x < 40
5
40 ≤ x < 50
7
50 ≤ x < 60
4
60 ≤ x ≤ 70
2
Draw a cumulative frequency diagram for this data
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Data Frequency
Cumulative Frequency
10 – 20 2 2
20 – 30 4 2 + 4 = 6
30 – 40 5 2 + 4 + 5 = 11
40 – 50 7 2 + 4 + 5 +7 = 18
50 – 60 4 2 + 4 + 5 +7 + 4 = 22
60 – 70 2 2 + 4 + 5 +7 + 4 + 2 = 24
Create a third CUMULATIVE FREQUENCY column like this
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You don’t have to show working
Data Frequency
Cumulative Frequency
10 – 20 2 2
20 – 30 4 2 + 4 = 6
30 – 40 5 2 + 4 + 5 = 11
40 – 50 7 2 + 4 + 5 +7 = 18
50 – 60 4 2 + 4 + 5 +7 + 4 = 22
60 – 70 2 2 + 4 + 5 +7 + 4 + 2 = 24
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You don’t have to show working
Data Frequency
Cumulative Frequency
10 – 20 2 2
20 – 30 4 6
30 – 40 5 11
40 – 50 7 18
50 – 60 4 22
60 – 70 2 24
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Data Frequency
Cumulative Frequency
10 – 20
2 2
20 – 30
4 6
30 – 40
5 11
40 – 50
7 18
50 – 60
4 22
60 – 70
2 24
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80
Data
Cunu
lative
Frequ
ency Series1
Plot these numbers
Plot the second number in the data column against the number in the cumulative frequency column
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Now, join up the points
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80
Data
Cum
ulat
ive
Freq
uenc
y
Series1
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You can use this to find the middle half
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80
¾ (18)
½(12)
¼(6) Lower Q
(30)Median
(42)
Upper Q(53)
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The Lower Quartile is 30Median is 42The Upper Quartile is 53
This means that the middle half is between 30 and 53. Called the inter-quartile range.53 – 30 = 23
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Frequency
Teach GCSE Maths
Grouped Data
Rainfall (mm) 2
718
17
5
1
and the Mean
<0 x < 2020 x < 3030 x < 35
40 x < 50
50 x < 70
35 x < 40
<
<
<
<
<
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e.g.1 This table gives the time taken for 30 components to fail.
Time to failure
(hours), t
Number of
compone
nts f
0 t < 20 5
20 t < 40 8
40 t < 60 17
<
<
<
means t can also equal 0.
Decide with your partner if t can equal 20 in the 1st class.
BUT, the extra line . . . 0 t < 20<
Tip: Tilt your head to the right and you can see the extra line making an equals
sign.Ans: No. Measurements of t = 20 are in the 2nd class.
Since the quantity is time, a t has been used instead of x.
The t written between 0 and 20 means that the time is between 0 and 20 hours !
The numbers 20, 40 and 60, at the top of the classes, are called the “upper class
boundaries”
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Tell your partner why, using the table, we
cannot find the exact value of the mean.
Suppose we want to find the mean time that a component lasts.
To calculate an estimate of the mean, we need to choose one number in each class that represents the class.
Ans: We don’t know the exact value of each time. For example, in the 1st class there are 5 failures. They could all have been in the 1st hour, or be equally spaced, or be 13·5, 16·2, 17, 18·7, 19·9 . . . or any times between 0 and 20.
Ans: t = 10. It is the mid-point of the class, the average of 0 and 20.
To represent a class, we use the mid-point of the class.
Time to failure
(hours), t
Number of
compone
nts f
0 t < 20 5
20 t < 40 8
40 t < 60 17
<
<
<
Decide with your partner which number you would use to represent the 1st class ( 0 t < 20 ).<
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We will need an extra column for the mid-points ( which can also be called t ).
Time to failure
(hours), t
Number of
compone
nts f
0 t < 20 5
20 t < 40 8
40 t < 60 17In this question, the mid-points are easy to spot but we need to remember that a mid-point is the average of the numbers at each end of the class ( the boundary values ).
(0 + 20) = 1012
(20 + 40) = 3012
(40 + 60) = 5012
<
<
<
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Time to failure
(hours), t
Number of
compone
nts f
Mid-point
0 t < 20 5
20 t < 40 8
40 t < 60 17
Time to failure
(hours), t
Number of
compone
nts f
Mid-point t × f
0 t < 20 5
20 t < 40 8
40 t < 60 17
Totals
Now we can calculate an estimate of the mean time.mean time = total time ÷ number of
components= sum of t × fsum of f
=
50
240
850
30 1140This column now gives t.
t10
30
50
<
<
<
114030 = 38
hoursCheck:38 is between 0 and 60.
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1. The table shows the lengths of 25 pieces of wood.
360 l < 90
650 l < 60
940 l < 50
430 l < 40
310 l < 30
Frequen
cy fLength
(cm) l
<
<
<
<
<
Exercise
(a)Calculate an estimate of the mean length.(b) Which is the modal class?
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Length (cm) l
Frequency f
10 l < 30 3
30 l < 40 4
40 l < 50 9
50 l < 60 6
60 l < 90 3
Total 25
Mid-value
20
35
45
55
75
60
140
405
330
225
l × fSolution:
Length (cm) l
Frequency f
10 l < 30 3
30 l < 40 4
40 l < 50 9
50 l < 60 6
60 l < 90 3
<
<
<
<
<
(a) mean length = total length ÷ number of pieces = sum of l × f
sum of f
= 116025 = 46·4
cm
Check:46·4 is between 10 and 90.(b) the modal class is 40 l <
50<
1160
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Changing the Subject of a formula
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Substituting
35
-3
-31
149
-24
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Same Sign Subtract
Solve 2x + y = 8
and 5x + y = 17
3x + 0 = 9
x = 3
Substitute x = 3 in
Check in (not used directly to find y)
5 x 3 + 2 = 17
1
2
2 1
1
2
-
2 x 3 + y = 8 so y = 2
x = 3 and y = 2
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Different Signs Add
Solve 3x + 2y = 8
x - 2y = 0
4x + 0 = 8 so x = 2
Substitute x = 2 in to find y
3 x 2 + 2y = 8 so 2y = 2 so y = 1
Check in 2 - 2 x 1 = 0
x = 2 and y = 1
1
2
1 2
1
2
+
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Different amounts of x and y
Solve x + 2y = 11
and 3x + y = 18
Need either same number of x’s or y’s so
gives 3x + 6y = 33
(SSS) 0 + 5y = 15 so y = 3
Sub y = 3 in
Check in 3 x 5 + 3 = 18
x = 5 and y = 3
1
2
31
3 2
1
2
x 3
-
x + 2 x 3 = 11 so x = 5
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Sometimes...• We need to multiply both equations
Solve 5x + 2y = 15
and 3x - 3y = 51
1
2
• We could do x 3 then 1 2 2x3 4We would then have two new equations &
which can be added to cancel out y as before
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Word problems
2. A fruit machine contains 200 coins. These are either 20p or 50p. The total value of the coins is £65.20• How many of each coin are in the machine?
yx be s50p' ofnumber and , be s20p' ofnumber Let
)1(200 yx
)2(6520 What is the value in pence of x 20p’s ?
yx 5020