Download Math 102 / Core 143 Final Exam Solutions: Probability and Statistics - Prof. Aaron J. Robe and more Exams Statistics in PDF only on Docsity! Solutions for Practice for FINAL EXAM Math 102 / Core 143 A. Robertson 1. We have three boxes with unknown numbers in each. From each box we draw, with re- placement, 1000 times. From Box I we want to investigate the sum; from Box II we want to investigate the average; from Box III we want to investigate the product. We can use the Normal curve to approximate the results for A) Box I only B) Box I and II only C) Box I and III only D) All three boxes B: Only the sum of draws (and hence the average) are approximately Normal. The Central Limit Theorem says nothing about products. 2. A student performs a hypothesis test comparing 115 samples against an established norm. She gets a test statistics of 2.14. Which of the following is correct at a significance level of 5%? A) The p-value is 1.6% so we conclude that our sample is di!erent from the established norm B) The p-value is 1.6% so we conclude that our sample is not di!erent from the established norm C) The p-value is 2.5% so we conclude that our sample is di!erent from the established norm D) The p-value is 2.5% so we conclude that our sample is not di!erent from the established norm A: We use the Normal curve to find the area to the right of 2.14 and get 1.6%. Since the p-value is less than 5%, we conclude that the di!erence is real, and unlikely to be due to chance. 3. Roll a fair 8-sided die 64 times. Find the 84th percentile for the number of threes rolled. A) 8.00 B) 9.25 C) 10.65 D) 12.80 C: The box is a 0-1 box with 1/8th ones and 7/8ths zeros. We draw 64 times. We have EV = 8 and SEsum = ! 64 ! (1/8)(7/8) " 2.65. We want the 84th percentile, which on the Normal chart correspronds to z = 1.00 (we want the area between #z and z to be 64% so that we have 84% to the left of z. Converting z = 1.00 out of standard units, we get 1.00 = X!82.65 which gives 10.65. 4. You have in your possession a biased coin which has probability of landing heads 1/3. i) Find the chance that at least one head appears on five consecutive tosses. A) 32.9% B) 50.0% C) 66.7% D) 86.8% D: We have a small number of tosses so we can’t use the Normal curve. Instead, we use the NOT rule. We don’t want all tosses to be tails. The chance of this occurring is (2/3)5 " .132 or 13.2%. Hence, the probability we want is 100%# 13.2% = 86.8%. ii) Find the chance of getting exactly 8 heads out of 10 tosses. A) 99.7% B) 30.0% C) 5.2% D) 0.3% D: We use the Binomial formula with n = 10, k = 8, and p = 1/3. We have " 10 8 # (1/3)8(2/3)2 " .003 or .3%. 5. A remarkable (ficticious) study result in Norway found a correlation between a person’s IQ and their height. The average IQ of the study group was 100 with a standard deviation of 15, while the average height of the study group was 66 inches, with a standard deviation of 6 inches. The remarkable results was that the researchers found a correlation coe"cient of r = .3. i) What do you predict for the height of a person in the study group with a 110 IQ? A) 79.3 in B) 72.0 in C) 68.9 in D) 67.2 in D: Let x be IQ and y be height. We want to predict y given x, so we use: y # ay = r(SDy/SDx)(x# ax). Hence, we have y # 66 = .3(6/15)(110# 100), which gives a predicted height of 67.2 in.
