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.4.1 Here are the data from Exercise 2.3.10 on the num-ber of virus-resistant bacteria in each of 10 aliquots: 14 14 15 26 13 16 21 20 15 13 (a) Determine the median and the quartiles. (b) Determine the interquartile range. (c) How large would an observati

Sagot :

MrRoyal

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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