Consider discretizing a continuous attribute whose values are listed below: 3, 4, 5, 10, 21, 32, 43, 44, 46, 52, 59, 67 Using equal-width partitioning and four bins, how many values are there in the first bin (the bin with small values)? A. 1 B. 2 C. 3 D. 4 E. 5 F. 6

D.) 4

Step-by-step explanation:

Given the following values :

3, 4, 5, 10, 21, 32, 43, 44, 46, 52, 59, 67

Binning means grouping data values into intervals.

Using equal bin width and number of bins equal 4

Evaluating the data:

Lowest value = 3; highest = 67

To obtain a 4 bins with equal width;

A bin size of 20 should be suitable and will accommodate all the listed values :

Bins frequency

1 - 20 4

21 - 40 2

41 - 60 5

61 - 80 1

Number of values in first bin = 4

There are 4 values in the first bin

The data entries are given as:

3, 4, 5, 10, 21, 32, 43, 44, 46, 52, 59, 67

Start by calculating the range:

Divide the range by 4 (i.e. the number of bins) to calculate the class width

So, the classes and their frequencies are:

The first bin (i.e. class 3 to 19) has a frequency of 4.

Hence, there are 4 values in the first bin

Read more about class width and intervals at:

link

answer:

a decrease of 32

step-by-step explanation:

160 - 20% = 32