Bluman A.G. Elementary Statistics- A Step By Step Approach

340 Chapter 6 The Normal Distributionc. Does it seem reasonable that one month could havea rainfall amount less than 30 inches? Yes, theprobability is slightly more than 40%.d. Does it seem reasonable that the mean of a sampleof 32 months could be less than 30 inches? It’spossible since the probability is about 9%.22. Systolic Blood Pressure Assume that the mean systolicblood pressure of normal adults is 120 millimeters ofmercury (mm Hg) and the standard deviation is 5.6.Assume the variable is normally distributed.a. If an individual is selected, find the probability thatthe individual’s pressure will be between 120 and121.8 mm Hg. 0.1255b. If a sample of 30 adults is randomly selected, findthe probability that the sample mean will bebetween 120 and 121.8 mm Hg. 0.4608c. Why is the answer to part a so much smaller thanthe answer to part b? Means are less variable thanindividual data.23. Cholesterol Content The average cholesterol contentof a certain brand of eggs is 215 milligrams, and thestandard deviation is 15 milligrams. Assume thevariable is normally distributed.a. If a single egg is selected, find the probabilitythat the cholesterol content will be greater than220 milligrams. 0.3707 (TI: 0.3694)b. If a sample of 25 eggs is selected, find theprobability that the mean of the sample will belarger than 220 milligrams. 0.0475 (TI: 0.04779)Source: Living Fit.24. Ages of Proofreaders At a large publishing company,the mean age of proofreaders is 36.2 years, and thestandard deviation is 3.7 years. Assume the variable isnormally distributed.a. If a proofreader from the company is randomlyselected, find the probability that his or her age willbe between 36 and 37.5 years. 0.1567b. If a random sample of 15 proofreaders is selected,find the probability that the mean age of theproofreaders in the sample will be between 36 and37.5 years. 0.496325. Weekly Income of Private Industry InformationWorkers The average weekly income of informationworkers in private industry is $777. If the standarddeviation is $77, what is the probability that a randomsample of 50 information workers will earn, on average,more than $800 per week? Do we need to assume anormal distribution? Explain.Source: World Almanac. 0.0174 No—the central limit theorem applies.Extending the ConceptsFor Exercises 26 and 27, check to see whether thecorrection factor should be used. If so, be sure to includeit in the calculations.26. Life Expectancies In a study of the life expectancy of500 people in a certain geographic region, the mean ageat death was 72.0 years, and the standard deviation was5.3 years. If a sample of 50 people from this region isselected, find the probability that the mean lifeexpectancy will be less than 70 years. 0.002527. Home Values A study of 800 homeowners in a certainarea showed that the average value of the homes was$82,000, and the standard deviation was $5000. If 50homes are for sale, find the probability that the mean ofthe values of these homes is greater than $83,500. 0.014328. Breaking Strength of Steel Cable The averagebreaking strength of a certain brand of steel cable is2000 pounds, with a standard deviation of 100 pounds.A sample of 20 cables is selected and tested. Find thesample mean that will cut off the upper 95% of allsamples of size 20 taken from the population. Assumethe variable is normally distributed. 1963.10 pounds29. The standard deviation of a variable is 15. If a sample of100 individuals is selected, compute the standard errorof the mean. What size sample is necessary to doublethe standard error of the mean? s X 1.5, n 2530. In Exercise 29, what size sample is needed to cut thestandard error of the mean in half? 4006–4 The Normal Approximation to the BinomialDistributionA normal distribution is often used to solve problems that involve the binomial distributionsince when n is large (say, 100), the calculations are too difficult to do by hand usingthe binomial distribution. Recall from Chapter 5 that a binomial distribution has the followingcharacteristics:1. There must be a fixed number of trials.2. The outcome of each trial must be independent.6–42