Re: ISI 2014 PEA discussion

Posted by Abhitesh on Feb 14, 2017; 6:40am
URL: http://discussion-forum.276.s1.nabble.com/ISI-2014-PEA-discussion-tp7603173p7603186.html

Q14. This is from mean value theorem. We can always find k in (1,2) such that f'(k) = 3a+b. This is not necessarily k.
Answer should be (D).

Q29. Probability distribution table for X
X      :  -2         -1             0           1              2
P(X) : 1/5       1/5            1/5        1/5           1/5
E(X)=0  SD(X) = \sqrt(2)
Probability distribution table for Y=|X|
Y      :  2         1             0           1              2
P(Y) : 1/5       1/5            1/5        1/5           1/5
E(Y)= 6/5  SD(V) = \sqrt(2)
Probability distribution table for XY
XY      :  -4         -1             0           1              4
P(XY) : 1/5       1/5            1/5        1/5           1/5
E(XY) = 0
Now, P(XY) \not= P(X)P(Y). Therefore X and Y are dependent.
r= [E(XY)-E(X)E(Y)]/SD(X).SD(Y)=0