Solve problems and show work in problems. #6 - Assignment Example

1. Compute the mean and variance of the following discrete probability distribution. x P(x) 2 .50 8 .30 10 .20 x P(x) 2 0.5 0 -3.4 5.780 8 0.3 2.42.62.028100.22.04.64.2325.412.040Mean, Variance, The mean is 5.4 and variance is 12.04.2. The Computer Systems Department has eight faculty, six of whom are tenured. Dr. Vonder, the chair, wants to establish a committee of three department faculty members to review the curriculum. If she selects the committee at random: a. What is the probability all members of the committee are tenured?

The total number of ways of selecting three tenured faculties from six tenured faculties is The total number of ways of selecting three faculties from eight faculties is Hence, the probability that all members of the committee are tenured isP(All Tenured) = b. What is the probability that at least one member is not tenured? (Hint: For this question, use the complement rule.)The probability that at least one member is not tenured isP(At least one is not Tenured) = 1 – P(All Tenured) = 1 – 0.3571 = 0.6429 or 9/143.

New Process, Inc., a large mail-order supplier of women’s fashions, advertises same-day service on every order. Recently, the movement of orders has not gone as planned, and there were a large number of complaints. Bud Owens, director of customer service, has completely redone the method of order handling. The goal is to have fewer than five unfilled orders on hand at the end of 95% of the working days. Frequent checks of the unfilled orders follow a Poisson distribution with a mean of two orders.

Has New Process, Inc. lived up to its internal goal? Cite evidence.Mean, λ = 2 orders per dayThe probability of x success is given by Below table summarizes the calculation for Poisson distribution.xP(x)00.135310.270720.270730.180440.090250.036160.012070.003480.0009The probability of less than five unfilled orders on hand isP(x < 5) = P(0) + P(1) + P(2) + P(3) + P(4) = 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 = 0.9473 (< 0.95)No, New Process, Inc. has not lived up to its internal goal. This is because there is a chance of 94.

73% of the working days of having fewer than five unfilled orders on hand at the end; however, the goal is to have fewer than five unfilled orders on hand at the end of 95% of the working days.4. Recent information published by the U.S. Environmental Protection Agency indicates that Honda is the manufacturer of four of the top nine vehicles in terms of fuel economy. a. Determine the probability distribution for the number of Hondas in a sample of three cars chosen from the top nine.Let x represents the number of Hondas in the sample of three cars chosen from the top nine.

The probability distribution for the number of Hondas in a sample of three cars chosen from the top nine, P(x) is given by Hypergeometric distribution with below parameters: Number of items in the population, N = 9 Sample size, n = 3 Number of success in population, s = 4The probability distribution is given below:xP(x)01231.00b. What is the likelihood that in the sample of three at least one Honda is included?P(x ≥ 1) = 1 – P(0) = 1 – 0.1190 = 0.8810The likelihood that in the sample of three at least one Honda is included is about 88.10%.5.

According to the “January theory,” if the stock market is up for the month of January, it will be up for the year. If it is down in January, it will be down for the year. According to an article in the Wall Street Journal, this theory held for 29 out of the last 34 years. Suppose there is no truth to this theory. What is the probability this could occur by chance? There is no truth to this theory, therefore, the probability that this theory holds for a year is p = 0.5. Given: x = 29 and n = 34The probability remains same for every year and all the years are independent to each other.

Thus, this is a case of binomial distribution. Therefore, the probability that this theory held for 29 out of the last 34 years is The probability this could occur by chance is about 0.0000162 (or 0.00162%).