Mean – 14.6
Variance = 11.84
We need to know how many ways there are to select 2 items from a group of 5 without replacement. Order does not matter.
I used the formula for a combination with no repetition. That’s because the order doesn’t matter.
C(5,2) = n! /r! (n – r)!
C(5,2) = 5! /2! (5 – 2)!
C(5,2) = 5!/(2! * 3!) = 10
There are two parameters for choosing a combinations rule. First, is it a combination or a permutation? A combination is one where the order doesn’t matter (like a hand of poker: 2-6-9-J-K is the same as K-J-9-6-2), and permutation is one where the order does matter (like a combination lock: the combination 23-16-8 is not the same as 8-23-16).
The second is whether or not there is replacement. In this case, the question stated that samples are to be taken without replacement.
Therefore, I used the formula for permutation without replacement. I didn’t want to include a sample like 9-9, nor did I want to include both 9-13 and 13-9.
Here are the sample means in the same order as above
Taking the mean of the sample means:
And the variance (again using VARP because it’s the entire population)
In Excel, there are two sets of formulas for variance and standard deviation. One set is if you’re taking a sample, another is if you’re using the whole population. In most cases you use the formula for a sample, but in this case, it was stated to be the entire population.
To use the formula, just enter:
Standard error of the mean is the square root of the variance:
sqrt(4.44) = 2.107
This is the central limit theorem.
This says that the sample mean approaches the population mean as the number of samples increases.
In this case, the sample mean is exactly the same as the population mean, because we took every possible sample of a certain size from the population.
This is explains why a best-of-7 series is more likely to be won by the better team than a best-of-3 series. As the sample size increases, dumb luck plays less of a role in the overall outcome.