Question 1134734: 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 50 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 by Theo(13342)   (Show Source):
You can put this solution on YOUR website! c(50,4) = 230,300
the formula used is the combination formula of c(n,x)
n is the size of the group that is being drawn from.
x is the size of the group that is being drawn.
c(n,x) = n! / (x! * (n-x)!)
c(50,4) = 50! / (4! * 46!)
this is equivalent to (50 * 49 * 48 * 47 * 46!) / (4! * 46!)
the 46! in the numerator and denominator cancel out and you are left with:
(50 * 49 * 48 * 47) / (4 * 3 * 2 * 1)
evaluate this to get 230,300.
each set of 4 accounts will be different from any other set of 4 accounts.
a much simpler example that can show you what i mean would be 2 accounts taken from 4 accounts.
the number of possible accounts in that case will be c(4,2) = 4! / (2! * 2!) = (4 * 3 * 2 * 1) / (2 * 1 * 2 * 1) = 6.
if the 4 accounts are a, b, c, and d, then the possible sets of 2 accounts will be:
a,b
a,c
a,d
b,c
b,d
c,d
each set of 2 will have at least 1 member that will not be in any other set of 2.
for example, there are no other sets that have a,b in them.
the other sets can have a, and they can have b, but they will not have a,b.
this makes each set a unique combination of members.
this formula works for all size sets, only it's way too difficult to show you all the possible combinations.
something like that would best be done by computer.
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