Question

Need Help on question 5. (C)

5. Consider the following data from the United Nations Office on Drugs and Crime regarding homicide rates per 100,000 people in the following rather haphazardly chosen countries.

Australia	1.0
Canada	1.4
China	1.0
Cuba	4.5
Czech Republic	1.0
Israel	2.0
Jamaica	52.6
Japan	0.4
Norway	0.6
Pakistan	7.6
Switzerland	0.7
United States of America	4.7

5a. What is the mean number homicides per 100,000 people in these countries?

5b. What is the median number of homicides per 100,000 people in these countries?

5c. Why are the mean and the median so different in this dataset? In this case, which is more indicative of most people’s experiences across these countries, the mean or the median and why?

Answer 1

The sample can be rearranged in increasing order as

0.4
0.6
0.7
1
1
1
1.4
2
4.5
4.7
7.6
52.6

a) The mean is calculated as

$\bar{x}=\frac{\sum x_{i}}{n}=\frac{0.4+0.6+......+52.6}{12}=6.4583$

b) Median is the middle value of the data set hence

Median =

$\frac{1+1.4}{2}=1.2$

c) The mean and median are different in the data set because there lies an outlier which affects the mean most in increasing its value, hence the mean and median are different and it is also a skewed data set.

Here since the data set is not symmetric and not approximately normal hence it is advised to use median as the central tendency variable.