Answer:
a) [tex] Median=\frac{8.7+9.7}{2}=9.2[/tex]
[tex] Q_1 = \frac{7.4+7.8}{2}= 7.6[/tex]
[tex] Q_3 = \frac{10.7+11.9}{2}= 11.3[/tex]
b) [tex] IQR = Q_3 - Q_1 = 11.3-7.6=3.7[/tex]
c) [tex] Lower= Q_1 -1.5 IQR = 7.6 -1.5*3.7=2.05[/tex]
[tex] Upper= Q_3 +1.5 IQR = 7.6 +1.5*3.7=13.15[/tex]
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:
[tex] Median=\frac{8.7+9.7}{2}=9.2[/tex]
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:
[tex] Q_1 = \frac{7.4+7.8}{2}= 7.6[/tex]
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:
[tex] Q_3 = \frac{10.7+11.9}{2}= 11.3[/tex]
Part b
The IQR is defined as [tex] IQR = Q_3 - Q_1 = 11.3-7.6=3.7[/tex]
Part c
We define the normal limits like this:
[tex] Lower= Q_1 -1.5 IQR = 7.6 -1.5*3.7=2.05[/tex]
[tex] Upper= Q_3 +1.5 IQR = 7.6 +1.5*3.7=13.15[/tex]
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.
