Question

20. Is the following data on incomes positively or negatively skewed? You do not need to calculate skewness, but you should justify your answer Data in thousands: 45, 43, 32, 23, 45, 43, 47, 39, 21, 90, 230. 21. (a) Find the weighted mean center of population, where cities' populations and coordinates are given as follows: City 3.3 1.1 5.5 3.7 1.1 4.3 3.4 1.2 2.4 1.1 Population 34,000 6,500 8,000 5,000 1,500 (b) Find the unweighted mean center, and comment on the differences between your two answers. (c) Find the Euclidean distances of each city to the weighted mean center of population Continued

I need a help on question #21 D & E only!

I figure weighted/unweighted mean already

Weighted mean is : (3.336, 3.483) / Unweighted mean is: (2.94, 2.48)

thanks in advance!

Answer 1

We enter data into MS-Excel.

D) calculate weighted standard distance is as

Here 2 co ordinates X and Y , and weight (w) is as Population. Then

First we calculate weighted mean as

$\bar{X_{w}} =\frac{\sum_{i=A}^{E} (X_{i} * w_{i})}{\sum_{i=A}^{E} w_{i}} = 3.33$

And

$\bar{Y_{w}} =\frac{\sum_{i=A}^{E} (Y_{i} * w_{i})}{\sum_{i=A}^{E} w_{i}} = 3.48$

Let we calculate Standard Distance is as

$SD_{w}=\sqrt{\frac{\sum_{i=A}^{E} w_{i} (X_{i} - \bar{X_{w}})^2}{\sum_{i=A}^{E} w_{i}} + \frac{\sum_{i=A}^{E} w_{i} (Y_{i} - \bar{Y_{w}})^2}{\sum_{i=A}^{E} w_{i}}}$

$SD_{w}=\sqrt{1.42 + 1.43} = 1.69$

Excel work is as

Let Relative standard deviation is as

RSD= 100*SD(w) / mean(distance)

here men(distance) = sqrt(6^2 + 6^2) = sqrt(72) = 8.48

Therefor RSD = 100*1.69/8.48 = 19.91