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Cumulative Frequency
How to draw a cumulative
frequency graph
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
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
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
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
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
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
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)
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
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
<
<
<
<
<
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”
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 ).<
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
<
<
<
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.
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?
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
Changing the Subject of a formula
Substituting
35
-3
-31
149
-24
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
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
+
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
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
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
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