Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

.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 :

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