Question

A sample of 11 individuals shows the following monthly incomes.

	Individual	Income ($)
	1	1,500
	2	2,000
	3	2,500
	4	4,000
	5	4,000
	6	2,500
	7	2,000
	8	4,000
	9	3,500
	10	3,000
	11	43,000

a.	What would be a representative measure of central location for the above data? Explain.
b.	Determine the mode.
c.	Determine the median.
d.	Determine the 60th percentile.
e.	Drop the income of individual number 11 and compute the standard deviation for the first 10 individuals.

Answer 1

a) What would be a representative measure of central location for the above data? Explain.

Since, the data set has an outlier : 11th value. Hence, mean might give an inappropriate estimate.

Therefore, median will be the most appropriate estimate.

b) Determine the mode.

The maximum occurring value is $4000. 3 times

Mode = 4000

c) Determine the median.

In order to compute the median, the data needs to be put into ascending order, as shown in the table below

Position	Income ($) (Asc. Order)
1	1500
2	2000
3	2000
4	2500
5	2500
6	3000
7	3500
8	4000
9	4000
10	4000
11	43000

Since the sample size n = 11 is odd, we have that (n+1)/2 = (11+1)/2 = 6 is an integer value, the median is computed directly by finding the value located at position 6th, which is

median = 3000

d) Determine the 60th percentile.

We need to compute the 60% percentile based on the data provided.

Position	Income ($) (Asc. Order)
1	1500
2	2000
3	2000
4	2500
5	2500
6	3000
7	3500
8	4000
9	4000
10	4000
11	43000

The next step is to compute the position (or rank) of the 60% percentile. The following is obtained:

Percentile Position =

Percentile Position =

Percentile Position = 7.2

Since the position found is not integer, the method of interpolation needs to be used. The 60% percentile is located between the values in the positions 7 and 8. Those values, based on the data organized in ascending order, are 3500 and 4000.

The value of 7.2 - 7 = 0.2 corresponds to the proportion of the distance between 3500 and 4000 where the percentile we are looking for is located at. In fact, we compute

= 3600

This completes the calculation and we conclude that = 3600.

e) Drop the income of individual number 11 and compute the standard deviation for the first 10 individuals.

The sample size is n = 10. The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	Income ($)	Income ($)2
	1500	2250000
	2000	4000000
	2500	6250000
	4000	16000000
	4000	16000000
	2500	6250000
	2000	4000000
	4000	16000000
	3500	12250000
	3000	9000000
Sum =	29000	92000000

The sample variance s^2s2 is

Therefore, the sample standard deviation s is

s = 888.8194