Mathematics, 03.03.2020 19:57 loki285

Past records indicate that the probability of online retail orders
that turn out to be fraudulent is 0.08. Suppose that, on a given
day, 20 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.
a. What are the mean and standard deviation of the number of online
retail orders that turn out to be fraudulent?
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?

Answers: 3

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Answer from: cofran

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The mean of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

p = 0.08, n = 20

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

E(X) = np = 20*0.08 = 1.6

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21$

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

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887$

18.87% 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?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282$

32.82% 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?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 1) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

From itens b and c

$P(X \leq 1) = 0.1887 + 0.3282 = 0.5169$

$P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831$

48.31% probability that two or more online retail orders will turn out to be fraudulent.

Answer from: Quest

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

Answer the following and be the brainiest the mistake in the pattern is that : a. the decimal point

Answer from: Quest

What is the question?

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