Question

Students with the test scores listed as follows: 29, 81, 68, 43, 65, 45, 69, 78, 81, 72, 88, 99, 95, 24, 77, 52. Partition them into four bins by (1) equal-frequency (equi-depth) method, (2) equal-width method, and (3) an even better method (such as clustering).

Answer 1

1)Equal-Frequency Method

Data set Provided Is:29,81,68,43,65,45,69,78,81,72,88,99,95,24,77,52

Write These Elements In Ascending Order

After Arranging Element In Ascending Order Data Elements Look Like :

24,29,43,45,52,65,68,69,72,77,78,81,81,88,95,99

Now To Find Number Of Elements(Frequency) In Each Partition

Frequency=(Total Number Of Elements) ÷ (Number Of Partitions)  

In Our Case Frequency = 16 ÷ 4 = 4 (16 divided by 4 gives us 4)

Now From Data set sorted in Ascending Order Keep On including 4 elements from Left Into Bin

Bin 1: 24, 29,43, 45

Bin 2: 52, 65, 68, 69

Bin 3: 72, 77, 78, 81

Bin 4: 81, 88, 95 ,99

Note:Here We Can See 81 Is Present In Both Bin 3 And Bin 4,because Provided Data-set Has Two Occurrences Of 81 And Aim Of This Approach Is To Create Bins Such That All Of Them Have Same Number Of Data-Elements.

2)Equal-Width Method

Dataset Provided Is:29,81,68,43,65,45,69,78,81,72,88,99,95,24,77,52

Maximum Element in Dataset(Max)= 99

Minimum Element In Dataset(Min)=24

Lets Say Number Of Bins To Be Created(N)=4 ...(Given Information)

For equal width method,Width Could Be Given By Formulae

Width=(max-min)/N

Width=(99-24)/4=18.75 Which Could Be Rounded Off To Nearest Integer Value 19

so Width In This Case=19

Bin Range

Bin 1: 29,24 [24-43)

Bin 2: 43 ,45 ,52   [43-62)

Bin 3: 68, 65, 69, 78, 72 ,77   [62-81)

Bin 4: 81, 81, 88, 99, 95 [81-+)

All The Element In Data-Set That Lies In The Range Specified Are Included In Corresponding Bin.

In Set Theory [x,y) means from x to y including 'x' but excluding 'y'.

3)Clustering(k-Means)

K-means clustering is a simple unsupervised learning algorithm that is used to solve clustering problems. It follows a simple procedure of classifying a given data set into a number of clusters, defined by the letter "N," which is fixed beforehand

Here We Have To Create Four Clusters N=4

Step 1:Create 4 clusters And Randomly Assign Data-Element Into The Cluster.

Data set Provided Is:29,81,68,43,65,45,69,78,81,72,88,99,95,24,77,52

Random Clusters Mean-Value Of Cluster Elements

Cluster 1: { 24,43,29,68 } (24+43+29+68) ÷ 4= 41

Cluster 2: { 81,81,77,69 } (81+81+77+69) ÷ 4 = 77

Cluster 3: { 88,95,99,65 } (88+95+99+65) ÷ 4 =86.75

Cluster 4: { 72,52,78,45 } (72+52+78+45) ÷ 4 =61.75

Form New Clusters By Moving Data Element From One Cluster To Another One If Its Distance From Its Own Mean Is More Than Its Distance From Some Other Cluster Mean Value

(example: Distance Of 68 present In Cluster 1 From Mean Value Of Cluster 1 Is: 68-41=27 Whereas Distance Of 68 From Mean Of Cluster 4 Is 68-61.75=6.25 As 68 Is More Closer To Mean Of Cluster 4 It Moves Into Cluster 4,Same Goes With Other Elements.And Finally We Would Get New Clusters As Shown Below)

Clusters Newly Calculated Mean

Cluster 1: { 24,29,43,45 } (24+29+43+45) ÷ 4 = 35.25

Cluster 2: { 72,77,78,81,81 } (72+77+78+81+81) ÷ 5= 77.8  

Cluster 3: { 88,95,99 } (88+95+99) ÷ 3 = 94

Cluster 4: { 52,65,68,69 } (52+65+68+69) ÷ 4= 63.5

Again Form New Clusters By Moving Data Element From One Cluster To Another One If Its Distance From Its Own Mean Is More Than Its Distance From Some Other Cluster Mean Value

Clusters Newly Calculated Mean

Cluster 1: { 24,29,43,45 }   (24+29+43+45) ÷ 4 = 35.25

Cluster 2: { 72,77,78,81,81 } (72+77+78+81+81) ÷ 5= 77.8

Cluster 3: { 88,95,99 } (85+95+99) ÷ 3 = 94

Cluster 4: { 52,65,68,69 }   (52+65+68+69) ÷ 4= 63.5

As None Of The Data Element Moved From One Cluster To Another Cluster Final Clusters Formed After Performing k-Means Clustering Are

Cluster 1: { 24,29,43,45 }

Cluster 2: { 72,77,78,81,81 }   

Cluster 3: { 88,95,99 }   

Cluster 4: { 52,65,68,69 }

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