At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Answer:
0.4629 = 46.29% probability that the manager finds more than two such transactions
Step-by-step explanation:
For each transaction, there are only two possible outcomes. Either there is a procedural error, or there is not. The probability of a procedural error on a transaction is independent of any other transaction. This means that we use the binomial probability distribution to solve this question.
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.
A manager randomly selected 25 large cash transactions at a bank that were made in January.
This means that [tex]n = 25[/tex]
The chance for a procedural error is 10%
This means that [tex]p = 0.1[/tex]
What is the probability that the manager finds more than two such transactions?
This is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718[/tex]
[tex]P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994[/tex]
[tex]P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0718 + 0.1994 + 0.2659 = 0.5371[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5371 = 0.4629[/tex]
0.4629 = 46.29% probability that the manager finds more than two such transactions
Using the binomial probability relation, the probability that more than 2 such transactions is made is 0.463
Given the Parameters :
- Sample size, n = 25
- Probability of success, p = 0.10
- q = 1 - p = 1 - 0.10 = 0.90
Using the relation :
- [tex] P(x = x) = nCx \times p^{x} * q^{n-x} [/tex]
Substituting the values into the relation :
P(X > 2) = P(X = 3) + P(X = 4) +... + P(X = 25)
Using a binomial probability calculator :
P(X > 2) = P(X = 3) + P(X = 4) +... + P(X = 25)
P(X > 2) = 0.226 + 0.138 +... +..
P(X > 130) = 0.463
The probability that more than 2 such transactions is made is 0.463
Learn more : https://brainly.com/question/15607519
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.