NEED HELP /W HW QUESTIONS. HE PROVIDED SOLUTIONS (listed below) BUT THE EXPLANATIONS ARE NOT MY CUP OF TEA. NEED MORE CLARITY AND DETAILS SO PLEASE PROVIDE AS MUCH AS POSSIBLE! THANK YOU

11.1 A study made by a staff officer of an armored division showed that in a random sample of n = 40 days the division had on the average 1,126 vehicles in operating condition. Given that o = 135 for such data, what can this officer assert with 95% confidence about the maximum error if he uses X = 1,126 as an estimate of the actual daily average number of vehicles that this armored division has in operating condition?11.3 A study of the annual growth of a certain kind of orchid showed that (under controlled conditions) a random sample of n = 40 of these orchids grew on theaverage 32.5 mm per year. Given that o = 3.2 mm for such data, what can one conclude with 99% confidence about the maximum error if > = 32.5 mm is used as an estimate of the true average annual growth of this kind of orchid (under the given controlled conditions)?11.8 Before bidding on a contract, a contractor wants to be 95% confident that he is in error by no more than 6 minutes when using the mean of a random sample to estimate the average time it takes for a certain kind of adobe brick to harden. How large a sample will he need if he can assume that o = 22 minutes for the time it takes for such brick to harden?11.10 Before purchasing a large shipment of ground pork, a sausage manufacturer wants to be "pretty sure" that his error does not exceed 0.25 ounce when using the mean of a random sample to estimate the actual fat content per pound. If the standard deviation of the fat content is known to be 0.77 ounce per pound, how many one-pound samples would he need if by "pretty sure" he meant 95% confident?11.14 With reference to Example 11.4, on page 267 where we had n = 150, = 69.5, and o = 6.2, suppose that we had asked for a 97% confidence interval. (a) Find the value of zoos from Table I. (b) Use the value of zoos obtained in part (a) to calculate a 97% z interval for the average mechanical aptitude of assembly-line workers in the given industry.11.20 Ten bearings manufactured by a certain process have a mean diameter of 0.406 cm with a standard deviation of 0.003 cm. Construct a 99% confidence interval for the mean diameter of bearings manufactured by this process. Assume that the population sampled is normal.11.22 Five containers of a commercial solvent, randomly selected from a large production lot, weigh 19.5, 19.3, 20.0, 19.0, and 19.7 pounds. Assuming that these data can be looked upon as a sample from a normal population, construct a 99% / interval for the mean weight of the containers of the solvent in the production lot.11.38 In a random sample, 64 of 200 motorists stopped at a roadblock had not fastened their seat belts. Construct a 95% confidence interval for the corresponding true proportion in the population sampled11.44 A random sample of 300 shoppers at a large supermarket includes 234 who regularly use discount coupons. Construct a 98% confidence interval for the probability that any one shopper at that supermarket, randomly chosen for an interview, will confirm that he or she regularly uses discount coupons.Question 11.1 The study has a sample n = 4-0. with sample mean f = 1.126. The population standard deviation 6 = 135. We want to know the maximum error if a? = 1,126 at 95% confidence. 0 The maximum error is calculated using the formula E = 3a}:2 '1\"? n 2e},2 = 2013250. in order to calculate the Z value,we need to find the 2 value that is corresponding to the probability 0.5 0.0250 2 0.4?50.'l'l1erefore,z value is1.96. E196135 4134 _' m" " Question 11.3 in a random sample of n = 40. sample mean f = 32.5. The population standard deviation 0 = 3.2. We want to know the maximum error if f = 32.5 at 99% confidence. 0 The maximum error is calculated using the formula E = Zn;2 - l Zeb = zmso. in order to calculate the 2 value. we need to Jl'ino' the 2 value that is corresponding to the probability 0.5 0.0050 2 0.4950.'l'l1erefore,z value is either 2.57 or 2.53. We will use 2.51 3.2 E = 2.5? - z 1.30. m Question 113 A contractor wants to he 95% confident that error = 6. The population standard deviation 0 = 22. We want to find out what should he the sampie size. 2 \".321 The sample size for estimating a: n = Zajz = 3013259. in order to calculate the Z uaine.we need to find the 3 value that is corresponding to the probability 0.5 9.0250 2 U.4?5D.Therefore.z uaine is 1.96. 11: 1.95 - 22 2 [] = 51.65. Therefore n = 52 after rounding up to the nearest integer. Question 11.10 Manufacturer wants error to not exceed 9.25 when using mean of a random sample. The standard deviation is 0.?! We want to know what sample to use in order to he 95% confident. 2 Eng-er] The sample size for estimating a: n = [ E 2a}:2 = 2013250. in order to calculate the Z vaine,we need to find the 2 value that is corresponding to the probability I15 0.0250 = 0.4?\"50. Therefore.z value is 1.96. [1.96 41?? n = 2 u 25 ] = 36.44. Therefore n = 37 after rounding up to the nearest integer. Question 11.14 n = 150, Y = 69.5,0 = 6.2. We are looking 97% confidence interval. (a)za/, = Zo.0150. In order to calculate the Z value, we need to find the z value that is corresponding to the probability 0.5 - 0.0150 = 0.4850. Therefore, z value is 2.17. (b) The formula for confidence interval is x - za/2 '