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Shkiper50 [21]
3 years ago
6

Past records indicate that the probability of online retail orders

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

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.

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)

So

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.

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Complete question:

A market analyst is hired to provide information on the type of customers who shop at a particular store. A random survey is taken of 100 shoppers at this store. Of these 100, 73 are women. The shoppers The data is summarized below. we grouped in three age categories, under 30, 30 up to 50 and 50 and over. Women under 30 are 30, Men under 30 are 8 Women 30 to 50 are 25 and Men are 14 Women 50 and over are 18 and Men are 5.

Let W be the event that a randomly selected shopper is a woman. Let A be the event that a randomly selected shopper is under 30. a.) Find the probability of W. W (women Shoppers) = Find the probability of A. b.) Find the probability of A and W. c.) P (A and W) (Shoppers) d.) Find the probability of A or W. P(A or W) (Shoppers) e.) Find the probability of A given W

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Step-by-step explanation:

- - - - - - - - Women Men Total

Under 30 - - - 30 - - - 8 - - 38

30 to 50 - - - - 25 - - 14 - - 39

50 & over - - - 18 - - - 5 - - -23

Totals - - - - - - 73 - - -27 - - 100

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Probability : P= (required outcome / Total possible outcomes)

A) probability of women shoppers: P(W)

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P(W) = 73/100

B) Probability of under 30: P(A)

= number of shoppers under 30 = 38

Tital number of shoppers = 100

P(A) = 38/ 100

C) probability of A and W; This is the probability that the selected person is a woman and under 30.

(W n A) = 30

= 30 / 100

D) probability of A or W:

P(A) + P(W) - P(A n W) :

(38 / 100 + 73 / 100 - 30 / 100) = (38+73-30) /100

= 81/100

E) probability of A given W:

P(A n W) / P(W) = 30/73

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Answer:

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