Stats HW-2

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Chapter 5: #10, 40, 48, 49, 66, 72 Chapter 6: #18, 34, 45, 49, 52, 53 China Stockbarger Chapter 5: 10. The percent frequency distributions of job satisfaction scores for a sample of information systems (IS) senior executives and middle managers are as follows. The scores range from a low of 1 (very dissatisfied) to a high of 5 (very satisfied). a. Develop a probability distribution for the job satisfaction score of a senior executive. x 1 2 3 4 5 f(x) .05 .09 .03 .42 .41 Mean = 4.05 b. Develop a probability distribution for the job satisfaction score of a middle manager. x 1 2 3 4 5 f(x) .04 .10 .12 .46 .28 c. What is the probability a senior executive will report a job satisfaction score of 4 or 5? (.42)+(.41)= .83 The probability that a senior executive will report a job satisfaction of 4 or 5 is .83 d. What is the probability a middle manager is very satisfied? The probability that a middle manager is very satisfied is .28 e. Compare the overall job satisfaction of senior executives and middle managers. Senior Executives are overall more satisfied because from 4 to 5 they are 83% satisfied whereas middle managers from 4 to 5 are 74% satisfied. A study conducted by the Pew Research Center showed that 75% of 18- to 34-year-olds living with their parents say they contribute to household expenses (The Wall Street Journal, October 22, 2012). Suppose that a random sample of fifteen 18- to 34-year-olds living with their parents is selected and asked if they contribute to household expenses. a. Is the selection of the fifteen 18- to 34-year-olds living with their parents a binomial experiment? Explain. Yes. Since the 18- to 34-year olds living with their parents are selected randomly, p is the same from trial to trial and the trials are independent. Binomial n= 15 and p= .75
b. If the sample shows that none of the fifteen 18- to 34-year-olds living with their parents contribute to household expenses, would you question the results of the Pew Research Study? Explain. The probability that none of the 15 living with their parents contributes to household expenses is .000 c. What is the probability that at least 10 of the fifteen 18- to 34-year-olds living with their parents contribute to household expenses? The probability that at least fifteen 18 to 34 year olds living with their parents contribute to household expenses is .8516. 48. In 2011, New York City had a total of 11,232 motor vehicle accidents that occurred on Monday through Friday between the hours of 3 p.m. and 6 p.m. (New York State Department of Motor Vehicles website, October 24, 2012). This corresponds to a mean of 14.4 accidents per hour. a. Compute the probability of no accidents in a 15-minute period. Family- Poisson Parameter: Mean= 14.4/ 4= 3.6 The probability of no accidents in 15 min time period is .02732 b. Compute the probability of at least one accident in a 15-minute period. The probability of at least one accident in a 15 mins period is .9727. c. Compute the probability of four or more accidents in a 15-minute period. The probability of four or more accidents in a 15 min period is .4848. 49. Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. a. Compute the probability of no arrivals in a one-minute period. The probability of no arrivals in a one minute period is .00004540. b. Compute the probability that three or fewer passengers arrive in a one-minute period. The probability that three or fewer passengers arrive in a one minute time period is .01034. c. Compute the probability of no arrivals in a 15-second period. Family= Poisson Mean= 2.5 The probability of no arrivals in a 15-second period is .08208. d. Compute the probability of at least one arrival in a 15-second period. The probability of at least one arrival in a 15-second period is .9179.
66. Many companies use a quality control technique called acceptance sampling to monitor incoming shipments of parts, raw materials, and so on. In the electronics industry, component parts are commonly shipped from suppliers in large lots. Inspection of a sample of n components can be viewed as the n trials of a binomial experiment. The outcome for each component tested (trial) will be that the component is classified as good or defective. Reynolds Electronics accepts a lot from a particular supplier if the defective components in the lot do not exceed 1%. Suppose a random sample of five items from a recent shipment is tested. a. Assume that 1% of the shipment is defective. Compute the probability that no items in the sample are defective. Binomial n= 5 p=.01 The probability that no items in the sample are defective is .9510 b. Assume that 1% of the shipment is defective. Compute the probability that exactly one item in the sample is defective. The probability that exactly one item in the sample is defective is .0480 c. What is the probability of observing one or more defective items in the sample if 1% of the shipment is defective? The probability of observing one or more defective items in the sample if 1% of the shipment is defective is .0490 d. Would you feel comfortable accepting the shipment if one item was found to be defective? Why or why not? Because the probability of one item was found to be defective is 5% I would be okay with knowing 95% of the time is fine. 72. Customer arrivals at a bank are random and independent; the probability of an arrival in any one-minute period is the same as the probability of an arrival in any other one-minute period. Answer the following questions, assuming a mean arrival rate of three customers per minute. a. What is the probability of exactly three arrivals in a one-minute period? The probability of exactly three arrivals in a one-minute period is .2240 b. What is the probability of at least three arrivals in a one-minute period? The probability of at least three arrivals in a one-minute period is .5768 Chapter 6: 18, 34, 45, 49, 52, 53 18. The average return for large-cap domestic stock funds over the three years 2009-2011 was 14.4% (AAII Journal, February, 2012). Assume the three-year returns were normally distributed across funds with a standard deviation of 4.4%.
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