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) = ⁿC_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) = ⁿC_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) = ⁿC_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