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Statistics - Test Error Rate











































Your company's internal auditor believes that 10% of the company's invoices contain errors. To check this theory, 20 invoices are randomly selected and 5 are found to have errors. What is the probability that of the 20 invoices written, five or more would contain errors if the theory is valid?

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Oct 11, 2011
Binomial Distribution - Test Error Rate
by: Staff

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Part II


You could compute the answer as shown above (by adding each of the terms together).

However, you can reduce the complexity of the calculations as follows:


The sum of all the possible probabilities = 1

1 = b(0; 20, 0.1) + b(1; 20, 0.1) + b(2; 20, 0.1) + b(3; 20, 0.1) + b(4; 20, 0.1) + b(5; 20, 0.1) + b(6; 20, 0.1) + b(7; 20, 0.1) + . . . + b(20; 20, 0.1)


These terms can be grouped:

1 = [b(0; 20, 0.1) + . . . + b(4; 20, 0.1)] + [b(5; 20, 0.1) + . . . + b(20; 20, 0.1)]


You are only interested in the following sum:

P(k ≥ 5) = [b(5; 20, 0.1) + . . . + b(20; 20, 0.1)]


You can calculate this sum by subtracting the first 5 terms from 1

[b(5; 20, 0.1) + . . . + b(20; 20, 0.1)] = 1 - [b(0; 20, 0.1) + . . . + b(4; 20, 0.1)]



P(k ≥ 5) = 1 - b(4; 20, 0.1) - b(3; 20, 0.1) - b(2; 20, 0.1) - b(1; 20, 0.1) - b(0; 20, 0.1)

P(k ≥ 5) = 1 - 20_C_4 * (0.1^4)*[0.9^(20-4)] - 20_C_3 * (0.1^3)*[0.9^(20-3)] - 20_C_2 * (0.1^2)*[0.9^(20-2)] - 20_C_1 * (0.1^1)*[0.9^(20-1)] - 20_C_0 * (0.1^0)*[0.9^(20-0)]


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20_C_4 = 20!/[4! * (20 - 4)!]
20_C_4 = 4845


20_C_3 = 20!/[3! * (20 - 3)!]
20_C_3 = 1140

20_C_2 = 20!/[2! * (20 - 2)!]
20_C_2 = 190

20_C_1 = 20!/[1! * (20 - 1)!]
20_C_1 = 20

20_C_0 = 20!/[0! * (20 - 0)!]
20_C_0 = 1



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P(k ≥ 5) = 1 - 4845* (0.9^4)*[0.1^(20-4)] - 1140 * (0.9^3)*[0.1^(20-3)] - 190 * (0.9^2)*[0.1^(20-2)] - 20* (0.9^1)*[0.1^(20-1)] - 1* (0.9^0)*[0.1^(20-0)]

P(k ≥ 5) = 0.0431744952845


The final answer to your question is: P(k ≥ 5) = 0.0431744952845




Thanks for writing.

Staff
www.solving-math-problems.com


Oct 11, 2011
Binomial Distribution - Test Error Rate
by: Staff

Part I


Question:

Your company's internal auditor believes that 10% of the company's invoices contain errors. To check this theory, 20 invoices are randomly selected and 5 are found to have errors. What is the probability that of the 20 invoices written, five or more would contain errors if the theory is valid?


Answer:

For this type of problem, you cannot use a continuous probability distribution.

You should use a discrete probability distribution such as the Binomial Distribution.

Solving this problem using the binomial distribution is shown below:

k: The number of invoices with errors.

n: The number of invoices selected.

P: The probability that an individual invoice has an error.

Q: The probability that an individual invoice does not have an error (Q = 1 - P).

b(k; n, P): Binomial probability - the probability that if n invoices are selected, exactly x invoices have errors, when the probability of an error on an individual invoice is P.

b(k; n, P) = n_C_k * (P^k)*[Q^(n-k)]

n_C_k: the binomial coefficient - how many ways can k be chosen from n?

n_C_k = n!/[k! * (n - k)!]



Binomial distribution for n = 20, p = 0.1


The probability that "exactly" k invoices out of a total of n invoices will have errors is::

b(k; n, P) = n_C_k * (P^k)*[Q^(n-k)]


The probability that "exactly" 5 invoices (k = 5) will have errors is::

b(k; n, P) = b(5; 20, 0.1) = 20_C_5 * (0.1^5)*[0.9^(20-5)]


The probability that "exactly" 6 invoices (k = 6) will have errors is::

b(k; n, P) = b(6; 20, 0.1) = 20_C_6 * (0.1^6)*[0.9^(20-6)]


The probability that "exactly" 7 invoices (k = 7) will have errors is::

b(k; n, P) = b(7; 20, 0.1) = 20_C_7 * (0.1^7)*[0.9^(20-7)]



The probability that 5 or more invoices (k = 5, 6, 7, ... 20) will have errors is::

P(k ≥ 5) = b(5; 20, 0.1) + b(6; 20, 0.1) + b(7; 20, 0.1) + . . . + b(20; 20, 0.1)
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