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Past insurance company audits have found that 2 percent of dependents claimed on an employee’s health insurance actually are ineligible for health benefits. An auditor examines a random sample of 7 claimed dependents. (a) What is the probability that all are eligible? (b) That at least one is ineligible?

Respuesta :

Using the binomial distribution, it is found that the probabilities are given by:

a) 0.8681 = 86.81%.

b) 0.1319 = 13.19%.

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.

In this problem, we have that:

  • 98% of the employees are eligible, hence p = 0.98.
  • A sample of 7 was taken, hence n = 7.

Item a:

The probability is P(X = 7), hence:

P(X = 7) = 0.98^7 = 0.8681 = 86.81%.

Item b:

The probability is:

P(X < 7) = 1 - P(X = 7) = 1 - 0.8681 = 0.1319 = 13.19%.

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

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