lesson 4 - linear coding

8/13/2019 Lesson 4 - Linear Coding

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Lesson 4 - Linear Coding



Linear Coding

Linear coding to find the mean

Linear Coding to find the standard deviation

Work References

ContentsContents



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



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



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



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



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



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



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




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




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




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



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



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 =



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



StandardDeviation


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



StandardDeviation


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



Linear Coding

Now complete Exercise 1F page 45

lesson 4 - linear coding

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