Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

AN ARTICLE STATED THAT “INTERNAL SURVEYS PAID FOR BY DIRECTORY ASSISTANCE PROVIDERS SHOW THAT EVEN THE MOST ACCURATE COMPANIES GIVE OUT WRONG NUMBERS 15% OF THE TIME.” ASSUME THAT YOU ARE TESTING SUCH A PROVIDER BY MAKING 10 REQUESTS AND ALSO ASSUME THAT THE PROVIDER GIVES THE WRONG TELEPHONE NUMBER 15% OF THE TIME. FIND THE PROBABILITY OF GETTING ONE WRONG NUMBER.

Sagot :

Using the binomial distribution, there is a 0.3474 = 34.74% probability of getting one wrong number.

What is the binomial distribution formula?

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

For this problem, the values of the parameters are given by:

p = 0.15, n = 10.

The probability of getting one wrong number is P(X = 1), hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

P(X = 1) = C(10,1) x (0.15)¹ x (0.85)^9 = 0.3474

0.3474 = 34.74% probability of getting one wrong number.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

#SPJ1