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Last_Singularity · #9 January 3rd, 2009, 07:05 AM

Quote:

Originally Posted by Bradley55

I think i remember dong a 1-var stats with something. is that possible with this method?

Your teacher has a lot of explaining to do, then, not us. Ask your teacher what she/he has taught you so far in class can be used to do this problem. I am out of ideas if she has not yet taught you the concepts of expected value or variance.

On the other hand, surely you understand the concept of expected value? It is simply a weighted average of all the possible values a variable $X$ can take on. For example, if you have a 50% chance of earned $100 and 50% chance of earning $200, then you EXPECT to earn, on average, 0.5(100)+0.5(200) = $150. In your case, you have four possible values for your variable instead of just two - just multiple each value by its probability and add all of such products up - that's your expected.

nzmathman did forget one thing: there is another way to calculate variance. Instead of using $var(x) = E(x^2) - (E(x))^2$ , we can use $var(x) = E((E(x)-x)^2)$

Using this method, variance is calculated by: $var(x) = \sum_{i=1}^n p(x_i) (\bar{x}- x_i)^2$ where the mean $\bar{x} = \sum_{i=1}^n p(x_i) x_i$

So, first figure out $\bar{x}$ : = $0 \times 0.4 + 1 \times 0.3 .....$ . You should get 1.

Then find the squared error of each $x_i$ from that mean and multiply those by their probabilities: $(0.4)(0-1)^2 + (0.3)(1-1)^2 + (0.2)(2-1)^2 + (0.1)(3-1)^2$ . That is your variance.

Remember to square root that to get standard deviation. You should get $1$ for both variance and standard deviation.