Question

.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

Answer 1

(a)

`Q_1 = 14`

`Median = 15`

`Q_3 = 20`

(b)
`IQR = 6`

Explanation:

Given

`14\ 14\ 15\ 26\ 13\ 16\ 21\ 20\ 15\ 13`

`n = 10`

Solving (a): Median and the quartiles

Start by sorting the data

`Sorted: 13\ 13\ 14\ 14\ 15\ 15\ 16\ 20\ 21\ 26`

The median position is:

`Median = (n + 1)/(2)`

`Median = (10 + 1)/(2) = (11)/(2) = 5.5th`

This implies that the median is the average of the 5th and the 6th data;

So;

`Median = (15+15)/(2) = (30)/(2) = 15`

Split the dataset into two halves to get the quartiles

`Lower: 13\ 13\ 14\ 14\ 15\`

`Upper: 15\ 16\ 20\ 21\ 26`

The quartiles are the middle items of each half.

So:

`Lower: 13\ 13\ 14\ 14\ 15\`

`Q_1 = 14` ---- 14 is the middle item

`Upper: 15\ 16\ 20\ 21\ 26`

`Q_3 = 20` ---- 20 is the middle item

Solving (b): The interquartile range (IQR)

This is calculated as:

`IQR = Q_3 - Q_1`

`IQR = 20 - 14`

`IQR = 6`

Solving (c): Incomplete details