Expected value of the population of a city in a region Given: A certain country has four regions: North, East, South, and West. The populations of these regions are 3 million, 4 million, 5 million, and 8 million, respectively. There are 4 cities in the North, 3 in the East, 2 in the South, and there is only 1 city in the West. Each person in the country lives in exactly one of these cities.
Part A: What is the average size of a city in the country?
My work: The probability of living in a north city is $\frac{4}{10}$, east city is $\frac{3}{10}$, south city is $\frac{2}{10}$, and west city is $\frac{1}{10}$.
So $E(X)=\sum xP(X)=3*\frac{4}{10}+4*\frac{3}{10}+5*\frac{2}{10}+8*\frac{1}{10}=\frac{21}{5}=4.2$ million people.
Part B: Show that without further information it is impossible to find the variance of the population of a city chosen uniformly at random. That is, the variance depends on how the people within each region are allocated between the cities in that region.
My best guess: Because variance would show the spread of the population from the average, we can't calculate it because we do not know the values of each individual city.
Part C: A region of the country is chosen uniformly at random, and then a city within that region is chosen uniformly at random. What is the expected population size of this randomly chosen city?
I am confused on how this would differ from part A. I'm given that the answer to C is larger than the answer to A though.
 A: The probability of living in a city in the North is $\frac {3\text {millon}}{20\text{million}}$
The average city size: There are 10 cities, and 20 million people.
$\frac {20\text{million}}{10}$
Estimating the variance, and why you need more information.
In the South, for example, there are $5$ million people between two cities.  These could be $1$ million and $4$ million or $100,000$ and $3.9$ million, or $2.5$ million and $2.5$ million.
Each will give a different variance number for exceptions of variance across the country.
Chebychev's inequality could give you and upper bound for the variance.
If a region is selected at random $\cdots$
The chance of any region being selected is $\frac 14$ times the average city size in each region.
$\frac 14 \frac 34 + \frac 14 \frac 43+\frac 14\frac 52+\frac 14 \frac {8}{1}$ million
A: There are $3+4+5+8=20$ million persons in the country.
There are $4+3+2+1=10$ cities in the country.
We conclude the cities of the country have two million persons on average.
