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pakinam

  • 2 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.) 1-Find E(X) 2-Find SE(X)

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  1. Khlara
    • 2 years ago
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    1)v1*p1+v2*p2+v3*p3+v4*p4

  2. kyusakazaki
    • 2 years ago
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    1.875

  3. abhi_abhi
    • 2 years ago
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    \[E(X)=\sum_{i=1}^{n}x _{i}p _{i} = 1*0.5+2*0.25+3*0.125+4*0.125 = 1.875\] \[SE(X)= \sqrt{E(X ^{2})-(E(X)})^2\] \[E(X ^{2}) =\sum_{i=1}^{n}x ^{2}_{i}p _{i} = 1^2*0.5+2^2*0.25+3^2*0.125+4^2*0.125 = 4.625\] \[SE(X)=\sqrt{E(X^2)-(E(X))^2} = \sqrt{4.625 -(1.875)^2} = 1.053268722\]

  4. JULIAKAPRI
    • 2 years ago
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    OK

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