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Sagot :
Answer:
(a)
[tex]Q_1 = 14[/tex]
[tex]Median = 15[/tex]
[tex]Q_3 = 20[/tex]
(b) [tex]IQR = 6[/tex]
Step-by-step explanation:
Given
[tex]14\ 14\ 15\ 26\ 13\ 16\ 21\ 20\ 15\ 13[/tex]
[tex]n = 10[/tex]
Solving (a): Median and the quartiles
Start by sorting the data
[tex]Sorted: 13\ 13\ 14\ 14\ 15\ 15\ 16\ 20\ 21\ 26[/tex]
The median position is:
[tex]Median = \frac{n + 1}{2}[/tex]
[tex]Median = \frac{10 + 1}{2} = \frac{11}{2} = 5.5th[/tex]
This implies that the median is the average of the 5th and the 6th data;
So;
[tex]Median = \frac{15+15}{2} = \frac{30}{2} = 15[/tex]
Split the dataset into two halves to get the quartiles
[tex]Lower: 13\ 13\ 14\ 14\ 15\[/tex]
[tex]Upper: 15\ 16\ 20\ 21\ 26[/tex]
The quartiles are the middle items of each half.
So:
[tex]Lower: 13\ 13\ 14\ 14\ 15\[/tex]
[tex]Q_1 = 14[/tex] ---- 14 is the middle item
[tex]Upper: 15\ 16\ 20\ 21\ 26[/tex]
[tex]Q_3 = 20[/tex] ---- 20 is the middle item
Solving (b): The interquartile range (IQR)
This is calculated as:
[tex]IQR = Q_3 - Q_1[/tex]
[tex]IQR = 20 - 14[/tex]
[tex]IQR = 6[/tex]
Solving (c): Incomplete details
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