Expected value of the population of a city in a region

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

a) There are 20 million total people across 10 cities, so 2 million per city.

b) Your intuition is correct. You don't know how many live in each city (except for the city in the West)

c) Instead of each city being weighted equally, each region is weighted equally. The average city size in North is 3 million / 4 cities, or $\frac{3}{4}$. We carry this calculation to see in N, E, S, W we have $\frac{3}{4}, \frac{4}{3}, \frac{5}{2}, \frac{8}{1}$ million people per city on average, respectively. Thus, if we were to select a region uniformly, we have

$\frac{1}{4} (\frac{3}{4} + \frac{4}{3} + \frac{5}{2} + \frac{8}{1}) = \frac{151}{48}$

Hope this helps!