miércoles, noviembre 12, 2003
[ 419. Stats ]
Except for this question. I'm still pondering over it.
Question 23. A random variable X is distributed X ~ B ( 5 , p ). It is given that Var(X) = 0.25 E(X). Find the value of E(X²).
My solution.
E(X) = np = 5p
Var(X) = npq = 5p(1-p)
thus,
20p(1-p)=5p
4p(1-p)=p
4p-4p²-p=0
p(3-4p)=0
p=0.75
since Var(X)=E(X²)-[E(X)]²
thus 5(0.75)(0.25)=E(X²)-[5(0.75)]²
0.9375=E(X²)-14.0625
E(X²)=15
Oh. only NOW when i work it out then i get 15. SHARKS!!!!
Oh for those who are interested, this question is worth 5 marks.
Shannon left at 1:44 p. m..
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