In order to verify the accuracy of their financial accounts, companies use auditors on a regular basis to verify accounting entries. The company's employees make erroneous entries 22% of the time. Suppose that an auditor randomly checks three entries. (a) Find the probability distribution for Y, the number of errors detected by the auditor.

P(Y=0)=

P(Y=1)=

P(Y=2)=

P(Y=3)=

(b)

Construct a probability histogram for p(y).

(c)

Find the probability that the auditor will detect more than one error.

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rarr solution.Let Y-denote the no of errors detected by auditorn=3p.prob. of exploy detecting error that makes emplayees =0.22Therefore,{:[Y∼" Binomial "(n=3","p=0.22)],[:.P(y=O/)=n//6p^(y)],[P(y=y)=^(n)C_(y)p^(y)(1-p)^(n-y)]:}Therefore,{:[p(y=0)=^(3)(_(0)(0.22)^(0)(1-0.22)^(3):}],[=0.4746],[p(y=1)=^(P)c_(1)**(0.22)^(1)(1-0.22)^(2)],[=0.40154],[=3C_(2)**(0.22)^(2)(1 ... See the full answer