Question

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.05. Suppose that, on a given day, 18 online retail orders are placed. Assume that the number of online retail orders that turn out to be fraudulent is distributed as a binomial random variable. Complete parts (a) through (d) below.

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent? The mean number of online retail orders that turn out to be fraudulent is nothing. (Type an integer or a decimal. Round to four decimal places as needed.) The standard deviation of the number of fraudulent retail orders is nothing. (Type an integer or a decimal. Round to three decimal places as needed.)

b. What is the probability that zero online retail orders will turn out to be fraudulent? nothing (Type an integer or a decimal. Round to four decimal places as needed.)

c. What is the probability that one online retail order will turn out to be fraudulent? nothing (Type an integer or a decimal. Round to four decimal places as needed.)

d. What is the probability that two or more online retail orders will turn out to be fraudulent? nothing (Type an integer or a decimal. Round to four decimal places as needed.)

Answer #1

Let X be the no. of fraudulent retail order.

n= 18 and p = 0.05

**a. Mean**=
18 * 0.05

**Ans: E(X) = 0.9**

**Variance**=
0.9 * 0.95 (q = 1 - p)

S.D. =

**Ans: S.D.(X) = 0.9247**

**b.** P (X = x) =

Zero frauds

P( X = 0) =

**Ans: Prob of zero frauds is 0.3972.**

**c.** One fraud

P ( X = 1) =

**Ans: Prob of one fraud is 0.3763.**

d. Probability of two or more frauds.

**
**

*Total probability is 1. The sum of an event and it's
complimentary is 1.
are complements. Hence, the above eq.*

= 1 - (0.3972 + 0.3763)

**Ans: Prob of two or more frauds is 0.2265.**

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