Given the following points (-2,0), (-1,0), (0,1), (1,1) and (2,3) What is the Y intercept of the regression line? a. 1.0 b. 1.5 c. 0.7 d. 0.0

Well normally you would need to work out the regression equation but here you are helped a lot by the fact that the regression line goes through (X-bar, Y-bar) and X-bar is conveniently 0 … Edit: So the answer is a = (3 + 1 + 1)/5

a.

You are right, but would like to know how you arrive at 3+1+1. Also, how would you solve this; A variable X is distributed normally and has a mean of 10. if the probability that an observation of X will be negative is 0.16, what is the Coefficient of Variation? a. 10.0 b. 1,0 c. 0.1 d. 0.32

(3+1+1+0+0) /5 …the average of Y coordinates

webtwister1 Wrote: ------------------------------------------------------- > You are right, but would like to know how you > arrive at 3+1+1. > That’s calculating Y-bar (the mean of the Y’s) so it’s actually, in order, 0 + 0 + 1 + 1 + 3 > Also, how would you solve this; > > A variable X is distributed normally and has a > mean of 10. if the probability that an observation > of X will be negative is 0.16, what is the > Coefficient of Variation? > > a. 10.0 > b. 1,0 > c. 0.1 > d. 0.32 P(X < 0) = 0.16 P(X - mu/sigma < -mu/sigma) = 0.16 P(Z < -10/sigma) = 0.16 Since P(Z < -1) = 0.16 then -10/sigma = -1 => sigma = 10 => CV = sigma/mu = 10/10 =1

Im not sure if this is the right way to look at the 1st question. BUt hey, isnt the 3rd point conveniently a (0,1)?? Which means that its your only y-intercept! I dont recall mention in schweser about divide by no of observations. If you happen to have two point on the y - axis, (0,y1); (0,y2), it makes sense for you to divide by the extremes. Problem is if there is no point on the y-axis proper… Joey, can u explain how to derive Sd from the probability? Thought it was quite ingenious about the part where you can form Z by minusin away mean divide by sigma. But how did u know -10/sigma = -1?

yea I started out along the same way and got stuck since I had this formula: (X-10)/sigma = -1 Not sure how Joey got -10/sigma = -1…help…

Can smeone explain how Joey got his sigma please?

bcos P(z

Right - from tables P(Z < -1) = 0.16 (or maybe that’s one you should know without tables) so if I have P(Z < [anything]) = 0.16 then [anything] must be -1. Here I had [anything] was -10/sigma so -10/sigma = -1 => sigma =10 by cross-multiplying.