Question

1. [40 pts] Consider a certain city with population 50000, where monthly income is normally distributed with mean 2000TL and standard deviation 300TL. (you may use the table below for calculations)
   1. [20pts] Find the maximum income among the people with the lowest 10% income.
   2. [20pts] Find the minimum income among the people with the highest 30% income.

Answer 1

A certain city with population 50000, has monthly income normally distributed with mean 2000 TL, and standard deviation of 300 TL.

So, if X is the montly income of a randomly selected individual from the city, then X follows normal with mean 2000 and standard deviation of 300.

So, Z = (X-2000)/300 follows standard normal with mean 0 amd standard deviation of 1.

Question a

We have to find the maximum income among the people, with the lowest 10% income.

So, we have to find m such that

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, we have

So, we conclude

So, the maximum income among the people with lowest 10% of income is 1616 TL.

Question b

We have to find the minimum income among the people with the highest 30% income.

So, we have to find n, such that

Where, phi is the distribution function of the standard normal variate.

So, we conclude that

So, the minimum income among the people with highest 30% income, is 2157.2 TL.