统计作业2003.doc

Chapter 5 Discrete Probability DistributionsP.20858. 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 has been tested. a. Assume that 1% of the shipment is defective. Compute the probability that no items in the sample are defective. P（0）=CnxPx（1-P）n-x=C50P0（1-P）5=C50*1*（1-1%）5=b. Assume that 1% of the shipment is defective. Compute the probability that exactly one item in the sample is defective. P（1）=CnxPx（1-P）n-x=C51P1（1-P）5-1=C51*1%*（1-1%）4=c. What is the probability of observing one or more defective items in the sample if 1% of the shipment is defective? P(X1)=1-P（0）=d. Would you feel comfortable accepting the shipment if one item were found to be defective? Why or why not?59. The unemployment rate is 4.1% (Barrons, September 4, 2000). Assume that 100 employable people are selected randomly.a. What is the expected number who are unemployed?Ex=n*p=100*4.1%=4.1b. What is the variance and standard deviation of the number who are unemployed? 2=np1-p=100*4.1%*(1-4.1%)= =np(1-p)=100*4.1%*1-4.1%=Chapter 6 Continuous Probability DistributionsP.23735. Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process (APICSThe Performance Advantage, July 1991). Assume a production process is designed to produce items with a weight of 10 ounces and that the process mean is 10. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situations.a. The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects.=10，=0.15z=X-=10.15-100.15=1 z=X-=9.85-100.15=-1 P(x10.15)=0.5-0.3413=0.1587P(xf9.85或x10.15)=2*(0.5-0.3413)=2*0.1587=0.31741000*0.3174=317.4b. Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. =10，=0.05z=X-=10.15-100.05=3 z=X-=9.85-100.05=-3 P(x10.15)=0.5-0.4987=0.0013P(xf9.85或x10.15)=2*(0.5-0.4987)=2*0.0013=0.00261000*0.0026=2.6c. What is the advantage of reducing process variation and setting process control limits at a greater number of standard deviations from the mean? It could reduce the probability of a defect.36. The mean hourly operating cost of a USAir 737 airplane is $2071 (The Tampa Tribune, February 17, 1995). Assume that the hourly operating cost for the airplane is normally distributed.a. If 11% of the hourly operating costs are $1800 or less, what is the standard deviation of hourly operating cost?Px1800=0.11,=2071 ,Z=NORMSINV0.11=-1.22653, =X-Z=1800-2071-1.22653=220.9485b. What is the probability that the hourly operating cost of a USAir 737 airplane is between $2000 and $2500?P2000x2071 , z=X-=2000-2071220.9485=-0.3213, P2000x2071=0.5-NORMSDIST-0.3213=0.1260P2071x2500, z=X-=2500-2071220.9485=1.9416, P2071x2500=NORMSDIST1.9416-0.5=0.4739P2000x2071+ P2071x2500=0.1260+0.4739=0.5999c. What is the hourly operating cost of the 3% of the airplanes that have the lowest operating cost? pxX=0.03, Z=NORMSINV0.03=-1.8807, X=Z*+=-1.88079*220.9485+2071=1655.442339. The Asian currency crisis of late 1997 and early 1998 was expected to lead to substantial job losses in the United States as inexpensive imports flooded markets. California was expected to be especially hard hit. The Economic Policy Institute estimated that the mean number of job losses in California would be 126,681 (St. Petersburg Times, January 24, 1998). Assume the number of jobs lost in California is normally distributed with a standard deviation of 30,000.a. What is the probability the number of lost jobs is between 80,000 and 150,000?Z1=80000-12668130000=-1.55603,Z2=150000-12668130000=0.7773 ,P(80000x1500000)= NORMSDIST0.7773-NORMSDIST-1.55603=0.781509-0.05985=0.721659b. What is the probability the number of lost jobs will be less than 50,000?Z=50000-12668130000=-2.55603,P(x50000)= NORMSDIST-2.55603=0.005294c. What cutoff value will provide a .95 probability that the number of lost jobs will not exceed the value?P(x)=0.95,Z=NORMSINV(0.95)=1.644854,X=Z*+=1.644854*30000+126681=176026.62P(X176026.62)=0.9542. A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must equal? Assume the filling weights have a normal distribution. =0.6, NORMSDIST(Z)=2%, Z=NORMSINV(0.02)=-2.05375, =X-Z*=18-(-2.05375*0.6)=3.03225Chapter 8 Interval EstimationP.30218. A USA Today study of rental car gasoline prices found the following prices per gallon at 12 major airports (USA Today, April 4, 2000). 1.58 1.53 1.60 1.55 1.80 1.75 1.58 1.62 1.69 1.21 1.50 1.55a. What is the point estimate of the population mean price per gallon?b. What is the point estimate of the population standard deviation?c. Assuming a normal population, what is the 95% confidence interval estimate of the population mean rental car price per gallon?P.31450. Mileage tests are conducted for a particular model of automobile. If the desired precision is a 98% confidence interval with a margin of error of 1 mile per gallon, how many automobiles should be used in the test? Assume that preliminary mileage tests indicate the standard deviation to be 2.6 miles per gallon.54. A USA Today/CNN/Gallup survey of 369 working parents found 200 who said they spend too little time with their children because of work commitments (USA Today, April 10, 1995).a. What is the point estimate of the proportion of the population of working parents who feel they spend too little time with their children because of work commitments?b. At 95% confidence, what is the margin of error?c. What is the 95% confidence interval estimate of the population proportion of working parents who feel they spend too little time with their children because of work commitments?58. A Roper Starch survey asked employees ages 18 to 29 if they would prefer better health insurance or a raise in salary (USA Today, September 5, 2000). Answer the following questions if 340 of 500 employees said they would prefer better health insurance over a raise.a. What is the point estimate of the proportion of employees ages 18 to 29 who would prefer better health insurance?b. What is the 95% confidence interval estimate of the population proportion?Chapter 9 Hypothesis TestingP. 37273. A bath soap manufacturing process is designed to produce a mean of 120 bars of soap per batch. Quantities over or under the standard are undesirable. A sample of 10 batches shows the following numbers of bars of soap. The population is assumed to be normal. 108 118 120 122 119 113 124 122 120 123Using a .05 level of significance, test to see whether the sample results indicate that the manufacturing process is functioning properly.75. Stout Electric Company operates a fleet of trucks that provide electrical service to the construction industry. Monthly mean maintenance cost has been $75 per truck. A random sample of 40 trucks provided a sample mean maintenance cost of $82.50 per month, with a sample standard deviation of $30. Managers want a test to determine whether the mean monthly maintenance cost has increased.a. Using a .05 level of significance, what is the rejection rule for this test?b. What is your conclusion based on the sample mean of $82.50?c. What is the p-value associated with this sample result? What is your conclusion based on the p-value?76. In making bids on building projects, Sonneborn Builders, I