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Sampling Distribution Practice Problems Solutions 1. The company JCrew advertises that 95% of its online orders ship within two working days. You select a random sample of 200 of the 10,000 orders received over the past month to audit. The audit reveals that 180 of these orders shipped on time. This is a categorical problem (sample proportion). a. What is the sample proportion of orders shipped on time? οΏ½ = 180/200 = 0.9 ππ b. Does the sample data satisfy conditions necessary for the sample proportion to follow an approximately normal distribution? np >= 5, 10, or 15 and n(1-p) >= 5, 10, or 15 οΏ½ = 0.9, p=0.95 (βrealβ number that ship on time according to JCrew) n=200, ππ (200)(0.95) = 190 (200)(1-0.95) = 10 Yes, conditions are satisfied c. What is the mean and standard error (SE) of the sample distribution assuming normal? mean = p = 0.95 SE = square root of [p(1-p)/n] = square root of [(0.95)(1-0.95)/200] = 0.015 d. If JCrew really ships 95% of its orders on time, what is probability that the proportion in a random sample of 200 orders is as small or smaller as the proportion in the audit? οΏ½ <= 0.9) when n=200. P(Z <=(ππ οΏ½ - p)/SE) = P(Z <=(0.9 - 0.95/0.015) = P(Z <= -3.33). Closest to P(ππ this on from the Standard Normal Table is -3.09, P = 0.001. Thus, P(Z <= -3.33) is less than 0.001. e. If we treated the problem as a binomial, how would the problem be set up? That is, what would we want to find the probability of? Trials would be 200, successes would be 180. P(X <= 180). 1 2. A Gallup poll found that for those Americans who have lost weight, 31% believed the most effective strategy involved exercise. What is the probability that from random sample of 300 Americans the sample proportion falls between 29% and 33%? Categorical (sample proportion). n = 300, p = 0.31 np = (300)(0.31) = 90 n(1-p) = (300)(1-0.31) = 210 Both >= 5, 10 or 15 οΏ½ <= 0.33) = P(ππ οΏ½ <= 0.33) β P(ππ οΏ½ <= 0.29) P(0.29 <= ππ SE = square root of [(0.31)(0.69)/300] = 0.0267 P(Z<= (0.33 β 0.31)/0.0267) β P(Z <= (0.29 β 0.31)/0.0267) P(Z<= 0.75) β P(Z <= - 0.75) From the Standard Normal Table, we find these two probabilities to be: 0.7734 β 0.2266 = 0.5468 3. A bottling company uses a machine to fill the bottles with olive oil. The bottles are designed to contain 475 milliliters (ml). In fact, the contents vary according to a normal distribution with a mean of 473 ml and standard deviation of 3 ml. Quantitative (sample mean). a. What is the distribution, mean, and standard error of the sample mean of six randomly selected bottles? Distribution is normal, sample mean should be normal with a mean = ΞΌ = 473, SE = sigma/square root of n = 3/β6 = 1.22. b. What is the probability that the mean of six bottles is less than 470 ml? οΏ½ < 470) = P(Z < (ππ οΏ½ β ΞΌ)/SE) = P(Z < (470-473)/1.22) = P(Z < - 0.246). From the Standard P(ππ Normal Table we find P(Z < - 0.246) = 0.0069. c. What is probability that the mean of six bottles is more than 475 ml? οΏ½ > 475) = P(Z > (ππ οΏ½ β ΞΌ)/SE) = P(Z > (475-473)/1.22) = P(Z > 1.64 ) = 1 β P(Z < 1.64). From the P(ππ Standard Normal Table for P(Z < 1.64) 0.9495 resulting in final probability of 1 β 0.9495 = 0.0505 2 4. Penn State Fleet which operates and manages car rentals for Penn State employees found that the tire lifetime for their vehicles has a mean of 50,000 miles and standard deviation of 3500 miles. Quantitative (sample mean). a. What would be the distribution, mean and standard error mean lifetime of a random sample of 50 vehicles? Mean = ΞΌ = 50,000, sigma = 3500. We canβt assume this is normal so have to go for the second assumption. Our sample will have to be larger than 30. Our sample is 50, so this works. n = 50 which is > 30, so we can assume sample mean to be approximately normal with mean = ΞΌ = 50,000 and SE = sigma/sr of n = 3500/β50 = 495. b. What is the probability that the sample mean lifetime for these 50 vehicles exceeds 52,000? οΏ½ > 52,000) = P(Z > (ππ οΏ½ β ΞΌ)/SE) = P(Z > (52,000 - 50,000)/495) = P(Z > 4.04). From the P(ππ Standard Normal Table we it only goes to 3.09, cumulative probability of 0.999. P(Z > 4.04) = 1 - 0.999 = 0.001. P(Z > 4.04) is less than 0.001. 3