Question #339648

A credit card company monitors cardholder transaction habits to detect any unusual activity. Suppose that the dollar value of unusual activity for a customer in a month follows a normal distribution with mean $250 and variance $391. a. What is the probability of $250 to $300 in unusual activity in a month? b. What is the probability of more than $300 in unusual activity in a month? c. Suppose that 10 customer accounts independently follow the same normal distribution. What is the probability that at least one of these customers exceeds $300 in unusual activity in a month? 


1
Expert's answer
2022-05-12T07:31:33-0400

a.


P(250<X<300)=P(Z<300250391))P(250<X<300)=P(Z<\dfrac{300-250}{\sqrt{391)}})

P(Z250250391))-P(Z\le\dfrac{250-250}{\sqrt{391)}})

0.99430.5=0.4943\approx0.9943-0.5=0.4943

b.


P(X>300)=1P(Z300250391))P(X>300)=1-P(Z\le\dfrac{300-250}{\sqrt{391)}})

0.0057\approx0.0057

c.

YBin(n,p)Y\sim Bin(n, p)


p=P(X>300)=0.0057,q=1p=0.9943,p=P(X>300)=0.0057, q=1-p=0.9943,

n=10n=10

P(Y1)=1P(Y=0)P(Y\ge 1)=1-P(Y=0)

=1(100)(0.0057)0(0.9943)100=1-\dbinom{10}{0}(0.0057)^0(0.9943)^{10-0}

0.055560\approx 0.055560


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