15. A financial consultant conducts seminars with each seminar limited to 20 attendees.
Because of the small size of the seminar, and the personal attention each person
receives, a large number of attendees are expected to become clients. From the past
experience, approximately 40% of the attendees become clients. What are the expected
value, variance, and the standard deviation of the number of clients becoming client.
(a) 6, 2.4, and 5
(b) 8, 5, and 1.5
(c) 40, 24, and 4.9
(d) 8, 4.8, and 2.2
(e) 8, 2.0, and 4.8
16. A production system with five production lines produces thousands of items per day.
From the past experience it is known that one out of every 10 products coming out of
each production line is defective. The probability of exactly two defective items in a
sample of 12 is:
(a) 0.2778
(b) 0.2530
(c) 0.2301
(d) 0.1933
(e) 0.2362
17. Which of the following statements about Poisson distribution are correct:
(a) The Poisson distribution is a discrete probability distribution
(b) The Poisson distribution is used in describing the number of events that occur in a
specified period of time
(c) The Poisson distribution can be used to calculate the probability of x number of
occurrences
(d) The statements (a), (b), and (c) are correct
(e) Only statements (a) and (b) are correct
18. Historical data indicate that 3 out of 100 products coming out of a production line are
defective. The quality inspector wants to select a sample of 100 items and would like to
determine the probability of certain number of products being defective or nondefective.
The appropriate distribution to find the probability of defective products from
a selected sample would be:
(a) Discrete probability distribution
(b) Binomial Distribution
(c) Normal distribution
(d) Poisson distribution
(e) Poisson’s Approximation
a
b
c
d
15. To find the expected value, variance, and standard deviation of the number of clients becoming clients, we need to use the formulas for a binomial distribution.
The expected value (μ) of a binomial distribution is given by: μ = n * p, where n is the number of trials and p is the probability of success.
In this case, n = 20 (number of seminar attendees) and p = 0.40 (probability of becoming a client).
So, μ = 20 * 0.40 = 8.
The variance (σ^2) of a binomial distribution is given by: σ^2 = n * p * (1 - p).
Plugging in the values, σ^2 = 20 * 0.40 * (1 - 0.40) = 4.8.
Finally, the standard deviation (σ) is the square root of the variance: σ = √(σ^2) = √4.8 ≈ 2.2.
Therefore, the answer is (d) 8, 4.8, and 2.2.
16. To find the probability of exactly two defective items in a sample of 12, we need to use the binomial probability formula.
The formula for the binomial probability is: P(X = k) = (n choose k) * p^k * (1 - p)^(n - k), where n is the number of trials, k is the number of successes, p is the probability of success, and (n choose k) is the binomial coefficient.
In this case, n = 12 (sample size), k = 2 (number of defective items), and p = 0.10 (probability of a product being defective).
Plug in the values to the formula:
P(X = 2) = (12 choose 2) * 0.10^2 * (1 - 0.10)^(12 - 2)
Calculate (12 choose 2) = 66:
P(X = 2) = 66 * 0.10^2 * (1 - 0.10)^(12 - 2)
P(X = 2) ≈ 66 * 0.01 * 0.9^10 ≈ 0.1933
Therefore, the answer is (d) 0.1933.
17. All of the statements (a), (b), and (c) are correct.
(a) The Poisson distribution is a discrete probability distribution.
(b) The Poisson distribution is used in describing the number of events that occur in a specified period of time.
(c) The Poisson distribution can be used to calculate the probability of x number of occurrences.
Therefore, the answer is (d) The statements (a), (b), and (c) are correct.
18. To find the probability of defective products from a selected sample, we need to use the appropriate distribution. Since we are dealing with a fixed number of trials (sample size) and a probability of success (defective product), the appropriate distribution is the binomial distribution.
Therefore, the answer is (b) Binomial Distribution.