Varience
Started by tgutberg
Look up standard deviation and variance. There are two formulas that you should probably know.
First, calculate the mean: (3+6+9+18)/4 = 9
Assuming this is asking for sample variance, the formula involved:
s = [(3-9)^2 + (6-9)^2 + (9-9)^2 + (18-9)^2]/(4-1)
= [36 + 9 + 0 + 81]/3 = 42
The exact question was:
The numbers (1,2,3,6) have an average of 3 and variance of 3.5. What is the average and variance of the set of numbers (3,6,9,18)?
the right answer is: 9, 31.5
where did you 31.5 come from? anyone got a clue?
The numbers (1,2,3,6) have an average of 3 and variance of 3.5. What is the average and variance of the set of numbers (3,6,9,18)?
the right answer is: 9, 31.5
where did you 31.5 come from? anyone got a clue?
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Damien was wrong in one thing: with variance, unlike st dev, you do not divide by the number of numbers-1(n-1), you just divide by 4. This is how you get 31.5
The mean is linear and the variance is quadratic (squared). Your new numbers are multiplied by 3. So the new mean is the old mean * 3 and the new variance is the old variance * 3^2 (or 9). And sure enough, 3.5*9 = 31.5.
And dividing by n or n-1 is not based on std dev or variance. You use the same for both. Std dev is just the square root of variance. You divide by n when you are dealing with a population. You use n-1 when you are using a sample because in a sample you have 1 less degree of freedom.
In this example they consider the set the entire population.
If it's a sample then you use n-1. If it's a population then you use n.
So if you have a population of a city of 10,000 people and you survey 50 of them, you would use n-1. If you surveyed all of them, you use n.
If they give you a set of numbers then those are the only numbers. I guess it's assumed you'd use just n. I doubt they'd have both answer choices down (one for n-1 and the other for n).
In any case, this problem wasn't testing your ability to determine variance. It was seeing if you knew that the mean is linear and the variance was quadratic in nature. You didn't even need the numbers, so long as it told you that all the numbers in the set were tripled in value.
So if you have a population of a city of 10,000 people and you survey 50 of them, you would use n-1. If you surveyed all of them, you use n.
If they give you a set of numbers then those are the only numbers. I guess it's assumed you'd use just n. I doubt they'd have both answer choices down (one for n-1 and the other for n).
In any case, this problem wasn't testing your ability to determine variance. It was seeing if you knew that the mean is linear and the variance was quadratic in nature. You didn't even need the numbers, so long as it told you that all the numbers in the set were tripled in value.
