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anonymous
 3 years ago
A random variable X has the probability distribution
value 1 2 3 4
probability 0.5 0.25 0.125 0.125
(For those of you who are interested, this is the geometric p=0.5 “killed” at 4. X is the number of times I toss a coin if I follow this rule: I’ll toss the coin till I get the first head, but I’ll stop after 4 tosses even if I haven’t got a head by that time.)
1Find E(X)
2Find SE(X)
anonymous
 3 years ago
A random variable X has the probability distribution value 1 2 3 4 probability 0.5 0.25 0.125 0.125 (For those of you who are interested, this is the geometric p=0.5 “killed” at 4. X is the number of times I toss a coin if I follow this rule: I’ll toss the coin till I get the first head, but I’ll stop after 4 tosses even if I haven’t got a head by that time.) 1Find E(X) 2Find SE(X)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01)v1*p1+v2*p2+v3*p3+v4*p4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[E(X)=\sum_{i=1}^{n}x _{i}p _{i} = 1*0.5+2*0.25+3*0.125+4*0.125 = 1.875\] \[SE(X)= \sqrt{E(X ^{2})(E(X)})^2\] \[E(X ^{2}) =\sum_{i=1}^{n}x ^{2}_{i}p _{i} = 1^2*0.5+2^2*0.25+3^2*0.125+4^2*0.125 = 4.625\] \[SE(X)=\sqrt{E(X^2)(E(X))^2} = \sqrt{4.625 (1.875)^2} = 1.053268722\]
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