Respuesta :
Answer:
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The mean of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]p = 0.08, n = 20[/tex]
a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?
Mean
[tex]E(X) = np = 20*0.08 = 1.6[/tex]
Standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21[/tex]
b. What is the probability that zero online retail orders will turn out to be fraudulent?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887[/tex]
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).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282[/tex]
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
[tex]P(X \leq 1) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
From itens b and c
[tex]P(X \leq 1) = 0.1887 + 0.3282 = 0.5169[/tex]
[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831[/tex]
48.31% probability that two or more online retail orders will turn out to be fraudulent.
The probability is an illustration of a binomial distribution.
The mean and the standard deviation
The given parameters are:
n = 20
p = 0.08
The mean is calculated as:
[tex]\bar x = np[/tex]
So, we have:
[tex]\bar x = 20 * 0.08[/tex]
[tex]\bar x = 1.6[/tex]
The standard deviation is calculated as:
[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]
This gives
[tex]\sigma = \sqrt{1.6 * (1 - 0.08)[/tex]
[tex]\sigma = 1.21[/tex]
Hence, the mean is 1.6 and the standard deviation is 1.21
The probability that zero online retail orders will turn out to be fraudulent
This is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(0) = ^{20}C_0 * 0.08^0 * (1 - 0.08)^{20 - 0}[/tex]
[tex]P(0) =0.1887[/tex]
The probability that zero online retail orders will turn out to be fraudulent is 0.1887
The probability that one online retail order will turn out to be fraudulent
This is calculated as:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(1) = ^{20}C_1 * 0.08^1 * (1 - 0.08)^{20 - 1}[/tex]
[tex]P(1) =0.3281[/tex]
The probability that one online retail orders will turn out to be fraudulent is 0.3281
The probability that two or more online retail orders will turn out to be fraudulent
This is calculated as:
[tex]P(x\ge 2) = 1 - P(0) - P(1)[/tex]
So, we have:
[tex]P(x\ge 2) = 1 - 0.1887 - 0.3281[/tex]
[tex]P(x\ge 2) = 0.4832[/tex]
The probability that two or more online retail orders will turn out to be fraudulent is 0.4832
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