Question

Question no 3: The average income of the residents of a particular community is your roll no *1000 and sd is your roll no *100:

What is the probability that income of a person, selected at random, is more than average income?

 What is the probability that income of a person, selected at random, is between average income and average income +2000?

What is the probability that income of a person, selected at random, is bbetween average income+1000 and average income +2000?

What is the probability that income of a person, selected at random, is less than average income +1000?

Answer 1

It is nesessary to asume, that roll no is the number given by a professor or a class instructor to the client. I will provide the common solution, but with no numbers in the answer as it will be impossible. To calculate the answer as a number, please refer to the Standart Normal Table: https://www.ztable.net/

Roll no will be named as $n$ later in the solution

Solution:

In this case, we can assume that the distribution is normal with mean equal to $1000n$ and standart deviation equal to $100n$.

1) We need to calculate the following probability:

$P\{\xi>1000n\}$

Next, we will use the following transformation and Standart Normal Table:

$P\{x_1\le\xi\le x_2\} = P\{(x_1-a)/\sigma \le (\xi-a)/\sigma \le (x_2-a)/\sigma\} =$

$= \Phi((x_2-a)/\sigma) - \Phi((x_1-a)/\sigma)$

$a = 1000n,$ $\sigma = \sqrt{100n} = 10\sqrt{n}$

Upper boundary is going to be $x_2 = +\infin$, $\Phi(+\infin) = 1$(adding or subtracting or divising will yeld no result), then let us calculate the lower boundary, or $x_1 = 1000n$:

$\Phi((1000n-1000n)/\sigma) = \Phi(0) = 0.5$

Finally, subtracting, we will get the answer:

$1 - 0.5 = 0.5$

2) We need to find the probability that income of a person is between average and average + 2000. It can be represented as:

$P(E(\xi)\le\xi\le E(\xi) + 2000) = P(1000n\leq \xi \le 1000n+2000)$

Using the transformation from 1):

$P(1000n\leq \xi \le 1000n+2000) = \Phi((1000n + 2000 - 1000n)/(10\sqrt n)) -$

$- \Phi((1000n-1000n)/(10\sqrt n)) = \Phi(2000/(10\sqrt n)) - \Phi(0) =$

$= \Phi(2000/(10\sqrt n)) - 0.5$

3) Following the same procedure as in 2) but for the different interval:

$P\{1000n +1000 \le \xi \le 1000n + 2000\} = \Phi((1000n + 2000 - 1000n)/(10\sqrt n)) -$

$\Phi((1000n + 1000)/10\sqrt n) = \Phi(2000/(10\sqrt n)) - \Phi(1000/ (10\sqrt n))$

4) Now we need to calculate the following probability:

$P(\xi < 1000n + 1000)$, for transformation lower boundary will be $- \infin$, $\Phi(-\infin) = 0$

So, the result will be:

$\Phi((1000n+1000-1000n)/(10\sqrt n)) = \Phi(1000/(10\sqrt n))$

Answer:

1) 0.5

2) $\Phi(2000/(10\sqrt n)) - 0.5$

3) $\Phi(2000/(10\sqrt n)) - \Phi(1000/ (10\sqrt n))$

4) $\Phi(1000/ (10\sqrt n))$