Answer to Question #339648 in Statistics and Probability for sezi

Question

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? 

Answer 1

a.

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

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

$$\approx0.9943-0.5=0.4943$$

b.

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

$$\approx0.0057$$

c.

$Y\sim Bin(n, p)$

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

$$n=10$$

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

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

$$\approx 0.055560$$