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1. The mean life of a computer disk drive is 2,000 hours. The standard deviation is 140 hours. Assuming the life-time of the drives to be normally distributed, find the probability of a disk drive lasting more than 1,800 hours. a. .4236 b. .9236 c. .8472 d. .5764 e. .2118 f. None of the above. z = (x – )/ = (1800 – 2000)/140 = -1.42857 P(x > 1800) = P(z > -1.42857) = 0.9236. 2. The yearly commissions per salesperson employed by a manufacturer of light machinery average $40,000 with a standard deviation of $5,000. What percent of the salespersons earn between $32,000 and $42,000? a. 60.06% b. 39.94% c. 34.13% d. 65.87% e. 81.66% f. None of the above. z = (x – )/ = (32000 – 40000)/5000 = -1.6 and z = (42000 – 40000)/5000 = 0.4 P(32000 < x < 42000) = P(-1.6 < z < 0.4) = 0.6006 This is 60.06%. 3. A study of 200 stamping firms revealed these incomes after taxes: Income After Taxes Number of Firms Under $1 million 102 $1 million up to $20 million $20 million and more 61 37 What is the probability that a particular firm selected has under $1 million in income after taxes? a. 0.000 b. 0.250 c. 0.333 d. 0.510 e. None of the above. P(Under 1 million) = 102/200 = 0.510. 4. A study of the opinion of designers with respect to the primary color most desirable for use in executive offices showed that: Primary Color Number of Opinions Red 92 Orange 86 Yellow 46 Green 91 Blue 37 Indigo 46 Violet 2 What is probability that a designer does not prefer red? a. 1.000 b. 0.770 c. 0.000 d. 0.230 e. None of the above. Total number of designers = 400 Number of designers who do not prefer red = 400 – 92 = 308 P(A designer does not prefer red) = 308/400 = 0.770 5.) The manufacturer of salad dressings uses machines to dispense the dressing into bottles that move along an assembly line. The machine is working well when 8 ounces is dispensed. The standard deviation of the process is 0.15 ounce. A sample of 50 bottles is selected periodically and the assembly line is stopped when there is evidence that the average amount dispensed is less than 8 ounces. Suppose that a sample of 50 bottles reveals an average of 7.983 ounces. 1. State the null and alternative hypotheses. 2. At the .05 level of significance, is there evidence that the average amount dispensed is less than 8 ounces? H0: The average amount dispensed is 8 ounces Ha: The average amount dispensed is less than 8 ounces Lower-tailed z- test for population mean with = 0.05 Lower Critical value of z = -1.645 Decision Rule: Reject H0 if z- score for the sample < -1.645 SE = s/n = 0.15/50 = 0.0212 z = (x-bar – )/SE = (7.983 – 8)/0.0212 = -0.8014 Since -0.8014 > -1.645, we fail to reject H0 There is no evidence that the average amount dispensed is less than 8 ounces. 6.) A machine being used for packaging raisins has been set so that, on average, 15 ounces of raisins will be packaged per box. The operations manager wishes to test the machine setting and selects a sample of 30 consecutive raisin packages filled during the production process. The sample mean is 15.18 and the sample standard deviation is 0.4909. Use the t-distribution. Is there evidence that the mean weight per box is different from 15 ounces? (use level of significance equal to .05) H0: The mean weight per box is 15 ounces Ha: The mean weight per box is not 15 ounces Two-tailed t- test for population mean with = 0.05 Degrees of freedom = n – 1 = 30 – 1 = 29; Critical value of t = 2.0452 Decision Rule: Reject H0 if |t- score| for the sample > 2.0452 SE = s/n = 0.4909/30 = 0.08963 t = (x-bar – )/SE = (15.18 – 15)/0.08963 = 2.008 Since 2.008 < 2.0452, we fail to reject H0 There is no evidence that the mean weight per box is different from 15 ounces.