Thread: Need help will Grade 12 statistics (weighted means, variance etc)

I am having a difficult time with understanding and using equations for weighted means, variance and deviation.
I have a question that lists the hours worked and frequency of workers in a table.
Hours Employees

12	1
32	5
35	7
38	8
42	5

I calculated the weighted mean (sum of all the hours times the frequency, divided by the total frequency).
I understand and correctly calculated 35.81 as the weighted mean. I'm not sure how to use this in order to find the standard deviation and variance.
I am struggling with it as a concept more so than I am struggling with numbers (if that makes sense).

*I may not understand certain math terms, as my learning has been messed up by moving to other countries, which required me to take earlier necessary courses, setting me back at least 1 grade level.

Hey Falkaine.

Basically the mean is calculated by multiplying each probability by its value and adding up the total. Notation wise we call this the expectation E[X].

When you calculate the frequency of one event and divide by the total number of events you get a probability and this weighted calculation you are doing is simply do this exact same thing (i.e. when you divide everything by the total and you see what that does to each term, it changes the frequency into a probability by normalizing the value).

Now the variance is calculated by using E([X-E[X]]^2) = E[X^2] - E[X]^2. Now E[X^2] is calculated by multiplying the probabilities by the square of the values instead of just the normal values.

So as an example, if we have a distribution with three values being 1,2,3 then E[X] = 1*P(X=1) + 2*P(X=2) + 3*P(X=3). But for E[X^2] this is 1^2*P(X=1) + 2^2*P(X=2) + 3^2*P(X=3).

Once you have the frequencies, you can calculate E[X] and E[X^2] very easily and you get the variance which is Var[X] = E[X^2] - {E[X]}^2 and the standard deviation is just the square root.

Intuitively, think about an expectation that weights things according to probability: if you have a higher probability it will attach more weight to that value than something that is smaller and what you do is you find a point (like a fulcrum or a see-saw) where half the weight is to the left and the other half is to the right where that one point balances out (like those old weight systems where to measure something, you put some weights on the right and an object on the left and you adjust the whole thing until it levels out).

Unbiased estimates of the mean and variance of a population, based on a random sample of 24 observations, are 5.5 and 2.42 respectively. Another random observation of 8.0 is obtained. Find new unbiased estimates of the mean and variance with this new information.
method plzz anyone plzzzzz reply fast but!!

The formula for the mean is x_bar = [X1 + X2 + X3 + X4 + .. + Xn]/n and the formula for sample variance (as opposed to normal variance is) [1/(n-1)]*[(X1-x_bar)^2 + (X2-x_bar)^2 + ... + (XN-x_bar)^2].

So you need to look at the the difference for these formulas for N observations and N+1 observations. I'll give you a hint with the first one:

Mean for N observations is [X1 + X2 + .. + XN]/n but for N+1 it's [X1 + X2 + ... + XN + X(N+1)]/(n+1). So if you have existing mean, multiply it by N add new observation and divide by N+1 you get:

New Mean = [(Old Mean)*N + New_Observation]/(N+1).

Can you do the same thing for the sample variance?

No thats where i need help bro help me and with values plzzzz

I will consult my horrible math teacher, I am really lacking in math vocabulary. Though your explanation seemed to be thorough, I simply didn't understand it. Evidently I am a preschooler in the context of math.

In expectations, you are not only restricted to finding the mean E[X]: you can also find the mean of a function of the random variable. So if you had say a random variable Y = X^2, then if you have the probabilities for X you can find E[Y].

This is required if you want to find the variance easily if you know E[X^2] and E[X]. Also if you have a function of a random variable (lets use the Y = X^2 example) then it means you can find the mean of Y without knowing it's PDF (but you know the PDF of X) and this is useful as well, since it's not always easy to have the PDF of the X but get the PDF of Y (for this example there is a way to do it easily, but not always in general).

So you can calculate E[X^3], E[SQRT(X)], E[e^(-X)] and so on, and there are lots of problems where you need to solve problems related to means of a transformed variable (like Y = X^2 or some other function), and the expectation formula tells us how to do this.