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Observe that for a random variable Y that takes on values 0 and​1, the expected value of Y is defined as​ follows:Upper E left parenthesis Upper Y right parenthesis equals 0 times Pr left parenthesis Upper Y equals 0 right parenthesis plus 1 times Pr left parenthesis Upper Y equals 1 right parenthesisE(Y) = 0×Pr(Y=0)+1×Pr(Y=1)​Now, suppose that X is a Bernoulli random variable with success probability Pr ​(X =1​)​= p. Use the information above to answer the following questions.Show thatUpper E left parenthesis Upper X cubed right parenthesis equals pEX3=p.Upper E left parenthesis Upper X cubed right parenthesisEX3​=​(00times×1 minus p1−p​)​+​(11times×​p)​= pp​(Usethe tool palette on the right to insert superscripts. Enter you answer in the same format asabove.​)Suppose that p​ =0.120.12.Compute the mean of X.Upper E left parenthesis Upper X right parenthesisE(X)​= nothing​(Roundyour response to two decimalplaces​)

1 Answer

6 votes

Answer:

0.074

Explanation:

X is Bernoulli random variable. Mean of X is given by p, wher p is the probability of success. Variance of X is given by p*(1-p)

p=0.08

So mean=0.08

Variance=0.08*0.92=0.074

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