Answer:
Interquartile range of the data is:
IQR = 2
Step-by-step explanation:
First lets rearrange the data in ascending order:
0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 4, 4
<h3>Find Median:</h3>
Median = (No. of terms +1) / 2 (when n = odd number)
Median = 14/2 (th term)
Median = 7th term
Median = 1
<h3>
Find Upper and Lower Medians:</h3>
As the number of total terms is odd (n=13) , we omit the median and form upper and lower sets and find their medium
0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 4, 4
Remove the median
0, 0, 0, 1, 1, 1 | 2, 2, 2, 3, 4, 4
Calculate the median of lower portion
0, 0, 0, 1, 1, 1
For even numbers
Lower Median = (0+1)/2
Lower Median = 0.5
Calculate the median of upper portion
2, 2, 2, 3, 4, 4
Upper Median = (2+3)/2
Upper Median = 2.5
<h3>
Interquartile Range:</h3>
It is the difference between Upper and Lower Median
IQR = 2.5 - 0.5
IQR = 2
terms of a geometric sequence is given by:
