Lesson 4 - Linear Coding
Transcript of Lesson 4 - Linear Coding
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Lesson 4 - Linear Coding
8/13/2019 Lesson 4 - Linear Coding
http://slidepdf.com/reader/full/lesson-4-linear-coding 2/19
Linear Coding
Linear coding to find the mean
Linear Coding to find the standard deviation
Work References
ContentsContents
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Linear Coding
Let’s begin by looking at the data listed below
4610 4612 4614 4616 4618 4620
Imagine that you were given the task of finding the mean of this set of numbers.The method that you would probably use would be as follows
mean = ( 4610 + 4612 + 4614 + 4616 + 4618 + 4620 ) = 27690 = 4615
6 6
Quite an easy task really, but now imagine that you were asked to do thecalculation in your head !
Unless you had an extremely good head for figures you would have to find away to try to simplify the problem
Mean
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Linear Coding
Let’s look at the data again x
4610
4612
4614
4616
4618
4620
Now we subtract 4610from each number
y = x - 4610
0
2
4
6
8
10
Now we can find the mean of these remainders by adding them together and
dividing by 6
mean = ( 0 + 2 + 4 + 6 + 8 + 10 ) = 30 = 5
6 6
Finally we add the 5 to the 4610 giving us a final mean of 4615 as before
Mean
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Linear Coding
This method works well if you can find a suitable code which will help you to simplifythe numbers. In this case y = x - 4610 was a suitable code but you could have chosen
another. For instance what would happen if we had chosen 4614 as the code ?
Well let’s see.
Now we need to subtract4614 from each number andas before find the mean of
the remainders
x
4610
4612
4614
4616
46184620
y = x - 4614
-4
-2
0
2
46
mean = ( -4 + -2 + 0 + 2 + 4 + 6 ) = 6 = 1
6 6
and then add together 4614 and 1 toonce again give a final mean of 4615
Mean
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Linear Coding
Now let’s look at a problem involving grouped data
The table below shows the times taken on 30 consecutive days for a coach tocomplete one journey on a particular route. We want to calculate the mean time fora journey using a method of coding.
Time (minutes) Frequencymid-points60 - 63
64 - 67
68 - 71
72 - 75
76 - 79
61.5
65.5
69.5
73.5
77.5
1
3
12
10
4
We now need to select a suitable code which will help us to calculate the mean
Mean
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Linear Coding
Time (minutes) Frequency(f)
mid-points
60 - 63
64 - 67
68 - 71
72 - 75
76 - 79
61.5
65.5
69.5
73.5
77.5
1
3
12
10
4
In this instance we are going to begin to create our code by selecting 69.5
and subtracting it from the mid-points
y = x - 69.5
-8
-4
0
4
8
And to further simplify the numbers we can divide by 4
making the code that we are going to use
y = x - 69.54
4
y = x - 69.5
-2
-1
0
1
2
Re-arranging this gives us x = 4y + 69.5
giving x = 4y + 69.5
Mean
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Linear Coding
Now from the table we find the following results
fy = 13
f = 30
giving y = 0.43
Now substituting this value for y into the equation we find
x = 4y + 69.5
x = 4 0.43 + 69.5 = 71.2
Hence the mean time for the journey was 71.2 minutes
Mean
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Linear Coding - Worked Example 1
The set of data tabulated below shows the weights (in gm) of a random selection ofletters. Using a method of coding calculate the mean weight of the letters
Weight (gm) Frequency (f)
304
308312
316
320
324
1
59
4
4
2
We will begin by selectingany one of the weights andsubtracting it from all theother weights to arrive at a
set of remainders
Let’s pick 316
Mean
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Linear Coding - Worked Example 1
Weight (gm) Frequency (f)
304
308
312
316
320
324
1
5
9
4
4
2
y = x - 316
-12
-8
-4
0
4
8
fy
-12
-40
-36
016
16
fy = -56
f = 25
giving y = -2.24 Hence mean x = 316 + y
= 316 - 2.24
= 313.76 gms.
Mean
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Linear Coding - Worked Example 2
In a Biology class students timed how long it took for a sample of their saliva to breakdown a 2% starch solution. The times to the nearest second are shown below. Using a
method of coding calculate the mean time.
Time (seconds) Frequency (f)
11 - 20
21 - 30
31 - 40
41 - 50
51 - 6061 - 70
71 - 80
1
2
5
11
82
1
Mid-points
15.5
25.5
35.5
45.5
55.565.5
75.5
Firstly we need to calculate the mid-points of the time data
Then we need to select a suitable mid-point. Let’s pick 45.5
Mean
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Linear Coding - Worked Example 2
Time (seconds) Frequency
(f)
11 - 20
21 - 30
31 - 40
41 - 50
51 - 60
61 - 70
71 - 80
1
2
5
11
8
2
1
mid-pts
15.5
25.5
35.5
45.5
55.5
65.5
80.5
y = x - 45.5
- 30
-20
-10
0
10
20
35
y = x - 45.5
10
-3
-2
-1
0
1
2
3.5
fy
-3
-4
-5
0
8
4
3.5
The coding to be used forthis problem is
y = x - 45.5
10
Mean
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Linear Coding StandardDeviation
As with the mean, standard deviation can also be foundsimply by extending your calculations
Fundamentally you will need to add to your table a calculationfor fy2.
Let’s look at some of the calculations that we made in previous examples andextend them to find a value for the standard deviation
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Linear Coding StandardDeviationThe table below shows the times taken on 30 consecutive days for a
coach to complete one journey on a particular route. We want tocalculate the standard deviation for the time taken for a journey
using a method of coding.
Time (minutes) Frequencymid-points
60 - 63
64 - 6768 - 71
72 - 75
76 - 79
61.5
65.569.5
73.5
77.5
1
312
10
4
We have already seen that we can find a value for themean by using the coding formula shown here 4
x - 69.5
Re-arranging this gives us x = 4y + 69.5 and sdx = 4sd y
y =
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StandardDeviation
Time (mins) Frequency
(f)
mid-pts
60 - 63
64 - 67
68 - 7172 - 75
76 - 79
61.5
65.5
69.573.5
77.5
1
3
1210
4
y = x - 69.5
-8
-4
04
8
y = x - 69.5
4-2
-1
01
2
fy fy2
-2
-3
010
8
4
3
010
16
Our table now needs to be extended to contain values for fy2
Now (sd y)2
= 33 - (0.43)2
= > sd y = 0.957
30
fy2 = 33
And sdx = 4 sd y
= 4 0.957
= 3.82 mins.
Linear Coding
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StandardDeviation
Linear Coding - Worked Example 1
In a Biology class students timed how long it took for a sample of their saliva to breakdown a 2% starch solution. The times to the nearest second are shown below. Using amethod of coding calculate the standard deviation for the time.
Time (secs) (f)
11 - 20
21 - 30
31 - 40
41 - 50
51 - 60
61 - 70
71 - 80
1
2
5
11
8
2
1
mid-pts
15.5
25.5
35.5
45.5
55.5
65.5
80.5
y = x - 45.5
- 30
-20
-10
0
10
20
35
y = x - 45.5
10
-3
-2
-1
0
1
2
3.5
fy
-3
-4
-5
0
8
4
3.5
fy2
9
8
5
0
8
8
12.25
fy2 = 50.25
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StandardDeviation
Linear Coding - Worked Example 1
Now (sd y)2 = 50.25 - (0.116)2 = > sd y = 1.29
30
And sdx = 10 x sd y
= 10 x 1.29
= 12.9 secs.
fy2 = 50.25
Now we know the following calculations
f = 30 Mean y = 0.116
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Linear Coding
Now complete Exercise 1F page 45