Question

Simple random sampling uses a sample of size n from a population of size N to obtain data that can be used to make inferences about the characteristics of a population. Suppose that, from a population of 80 bank accounts, we want to take a random sample of four accounts in order to learn about the population. How many different random samples of four accounts are possible?

Answer 1

Answer:

1,581,580

Explanation:

Here we have a combination problem, because the order of the selected accounts is not important. (if the order was important, it would be a permutations problem)

We have 80 possible accounts (n)

Out of which we need to select 4 (r)

C(n,r) = n! / (r! (n - r)!)

C(80,4) = 80! / (4! 76!) = 1,581,580

So, there are 1,581,580 different random samples of 4 accounts out of 80.

Answer 2

Different random samples of four accounts are possible, 1,581,580 when Simple random sampling uses a sample of size n from a population of size N.

What is random sampling?

Random sampling is defined as the component of the sampling method in which all sample has an equal probability of being picked out.

A sample selected at random is meant to be an impartial content of the total population.

In the above case, we have a combination trouble, because the order of the chosen reports is not crucial.

According to the given problem,

Possible accounts (n) =80

Selection(r) = 4

The formula of combination is :

$\text{C(n,r)} = \dfrac{ n! }{r! (n - r)!)}$

Substitute the given values in the above formula:

$\text{C(n,r)} = \dfrac{ n! }{r! (n - r)!)}\\\\C(80,4) =\dfrac{80!}{4!\times76!}\\\\C(80,4) =1,581,580$

Therefore, there are 1,581,580 different random samples of 4 accounts out of 80.

To learn more about the random sample, refer to:

https://brainly.com/question/12719654