At a financial institution, a fraud detection system identifies suspicious transactions and sends them to a specialist for review. The specialist reviews the transaction, the customer profile, and past history. If there is sufficient evidence of fraud, the transaction is blocked. Based on past history, the specialist blocks 40 percent of the suspicious transactions. Assume a suspicious transaction is independent of other suspicious transactions.

(a) Suppose the specialist will review 136 suspicious transactions in one day. What is the expected number of blocked transactions by the specialist? Show your work.

(b) Suppose the specialist wants to know the number of suspicious transactions that will need to be reviewed until reaching the first transaction that will be blocked.

(i) Define the random variable of interest and state how the variable is distributed.

(ii) Determine the expected value of the random variable and interpret the expected value in context.

(c) Consider a batch of 10 randomly selected suspicious transactions. Suppose the specialist wants to know the probability that 2 of the transactions will be blocked.

(i) Define the random variable of interest and state how the variable is distributed.

(ii) Find the probability that 2 transactions in the batch will be blocked. Show your work.

To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

ii. E(X) = 1/p , according to the geometric model.

= 1/.4 = 2.5.

We expect that the specialist will review 2.5 suspicious transactions on average before finding the first transmission that will be blocked.

c.

i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

= 45 (.4^2)(.6^10-2) = .0016

As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

aksnkj aksnkj · Answer 2 · 2021-10-10T17:09:04+02:00

Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = 0.0116

Step-by-step explanation:

a.

Here:

n:- number of trials

p:- probability

So,

[tex]E(x) = 0.40 x 136\\ = 54.4[/tex]blocked transmissions

To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

ii.

, according to the geometric model.

[tex]=\frac{1}{0.4} =2.5[/tex]

We expect that the specialist will review 2.5 suspicious transactions on average before finding the first transmission that will be blocked.

c.

i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

ii. P(Y=k) = (n choose k)()()

P(Y=2) = (¹⁰₂)

= 45 (.4^2)(.6^10-2) = .0016

As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

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