Question

What is the age distribution of patients who make office visits to a doctor or nurse? The following table is based on information taken from a medical journal. | Age group, years Under 15 15-24 25-44 45-64 65 and older | Percent of office visitors 10% 5% 15% 10% 60% Suppose you are a district manager of a health management organization (HMO) that is monitoring the office of a local doctor or nurse in general family practice. This morning the office you are monitoring has eight office visits on the schedule. What is the probability of the following? (a) At least half the patients are under 15 years old. (Round your answer to three decimal places.) Explain how this can be modeled as a binomial distribution with 8 trials, where success is visitor age is under 15 years old and the probability of success is 10%? Let n = 8, p = 0.15 and compute the probabilities using the binomial distribution. Let n = 8, p = 0.90 and compute the probabilities using the binomial distribution. Let n = 15, p = 0.10 and compute the probabilities using the binomial distribution. Let n = 8, p = 0.10 and compute the probabilities using the binomial distribution. (b) From 2 to 5 patients are 65 years old or older (include 2 and 5). (Round your answer to three decimal places.) (c) From 2 to 5 patients are 45 years old or older (include 2 and 5). (Hint: Success if 45 or older. Use the table to compute the probability of success on a single trial. Round your answer to three decimal places.) (d) All the patients are under 25 years of age. (Round your answer to three decimal places.) (e) All the patients are 15 years old or older. (Round your answer to three decimal places.)

Answer 1

a) n = 8

     p = 0.1

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 4) = P(x = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 7) + P(X = 8)

              = 8C4 * (0.1)^4 * (0.9)^4 + 8C5 * (0.1)^5 * (0.9)^3 + 8C6 * (0.1)^6 * (0.9)^2 + 8C7 * (0.1)^7 * (0.9)^1 + 8C8 * (0.1)^8 * (0.9)^0 = 0.005

Option - D) Let n = 8, p = 0.10 and compute the probabilities using the binomial distribution.

b) n = 8

   p = 0.6

P(2 < X < 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

                    = 8C2 * (0.6)^2 * (0.4)^6 + 8C3 * (0.6)^3 * (0.4)^5 + 8C4 * (0.6)^4 * (0.4)^4 + 8C5 * (0.6)^5 * (0.4)^3

                    = 0.676

c) n = 8

    p = 0.1 + 0.6 = 0.7

P(2 < X < 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

                    = 8C2 * (0.7)^2 * (0.3)^6 + 8C3 * (0.7)^3 * (0.3)^5 + 8C4 * (0.7)^4 * (0.3)^4 + 8C5 * (0.7)^5 * (0.3)^3

                    = 0.447

d) n = 8

    p = 0.1 + 0.05 = 0.15

P(X = 8) = 8C8 * (0.15)^8 * (0.85)^0 = 0.000

e) n = 8

p = 0.05 + 0.15 + 0.1 + 0.6 = 0.9

P(X = 8) = 8C8 * (0.9)^8 * (0.1)^0 = 0.430