Question

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.07. 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 ______.

b. What is the probability that zero online retail orders will turn out to be fraudulent?

c. What is the probability that one online retail order will turn out to be fraudulent?

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

If you use Excel can you show the formula used to get d? Thanks!

Answer #1

**Solution:-**

**a) The mean and standard deviation of the number of
online retail orders that turn out to be fraudulent is 1.26 and
1.0825.**

E(x) = n*p

E(x) = 18*0.07

E(x) = 1.26

S.D = sqrt(n*p(1-p)

**S.D = 1.0825**

**b) The probability that zero online retail orders will
turn out to be fraudulent is 0.2708 .**

x = 0

By applying binomial distribution:-

P(x,n) =
^{n}C_{x}*p^{x}*(1-p)^{(n-x)}

**P(x = 0) = 0.2708**

**c) The probability that one online retail order will
turn out to be fraudulent is 0.3669**

x = 1

By applying binomial distribution:-

P(x,n) =
^{n}C_{x}*p^{x}*(1-p)^{(n-x)}

**P(x = 1) = 0.3669**

**d) The probability that two or more online retail orders
will turn out to be fraudulent is 0.3622.**

x = 2

By applying binomial distribution:-

P(x,n) =
^{n}C_{x}*p^{x}*(1-p)^{(n-x)}

P(x > 2) = 1 - [P(x = 0) + P(x = 1)]

P(x > 2) = 1 - [0.2708 + 0.3669]

**P(x >
2) = 0.3622**

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