Here are the 18 measurements of MAO activity reported in Exercise 2.2.2: 6.8 8.4 8.7 11.9 14.2 18.8 9.9 4.1 9.7 12.7 5.2 7.8 7.8 7.4 7.3 10.6 14.5 10.7 (a) Determine the median and the quartiles. (b) Determine the interquartile range. (c) How large would an observation in this data set have to be in order to be an outlier? (d) Construct a boxplot of the data

a)

b)

c)

So on this case if a value is higher than 13.15 could be considered as an outlier

d) Figure attached

Step-by-step explanation:

For this case w ehave the following dataset:

6.8 8.4 8.7 11.9 14.2 18.8 9.9 4.1 9.7 12.7 5.2 7.8 7.8 7.4 7.3 10.6 14.5 10.7

Part a

The first step on this case is order the data on increasing way and we got:

4.1  5.2  6.8  7.3  7.4  7.8  7.8  8.4  8.7  9.7  9.9 10.6 10.7 11.9 12.7 14.2 14.5  18.8

Since we have 18 values the median can be calculated as the average of the 9th and 10th position of the dataset ordered and we got:

For the Q1 we can use the following dataset: 4.1  5.2  6.8  7.3  7.4  7.8  7.8  8.4  8.7  9.7 and Q1 would be given by:

For the Q3 we can use the following dataset: 8.7  9.7  9.9 10.6 10.7 11.9 12.7 14.2 14.5  18.8 and Q3 would be given by:

Part b

The IQR is defined as

Part c

We define the normal limits like this:

So on this case if a value is higher than 13.15 could be considered as an outlier

Part d

For this case we can use the following R code:

> x<-c(6.8, 8.4, 8.7, 11.9, 14.2, 18.8, 9.9, 4.1, 9.7, 12.7, 5.2, 7.8, 7.8, 7.4, 7.3, 10.6, 14.5, 10.7)

> boxplot(x, main="Boxplot of data")

The result is on the figure attached. As we can see the circle value 14.5 is represented as an outlier by the boxplot.

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11