1. (20 points) There is a continuous random variable X with cumulative distribution function: x < 0, 0 3 F (x) = 23 (x2 x3 ) if 0 x < 1, 1 x 1. (a) (4 points) Compute P(0.23 < X < 0.71). 3 F(0.71) - F(0.23) = 23 (0.712 0.71 ) 23 (0.232 3 0.233 ) 3 = 0.5772 0.0733 = 0.5039 (b) (4 points) Find the probability density function f (x). 3 2 d 3 d F (x) = dx (x2 x3 ) = 32 (2x x2 ) = 3x 3x2 0 < x < 1 f (x) = dx 2 (c) (4 points) Compute E[X]. R1 R1 2 x(3x 3x2 )dx = 0 3x2 0 3x3 dx 2 (d) (4 points) Compute E[X 2 ] R1 2 R1 3 3x2 x (3x )dx = 3x 2 0 0 = x3 3x4 dx 2 = 3 4 x 4 1 3x4 8 =1 0 1 3x5 10 = 0 3 4 3 8 = 5/8 = 0.625 3 10 = 0.45 (e) (4 points) Compute E[3 2.5X]. 3 2.5E[X] = 3 (2.5 0.625) = 1.4375 2. (20 points) The typical monthly smartphone bill for UCSB students follows a continuous uniform distribution with minimum of $85 and maximum of $145. The mean of monthly bill is $115 and variance is $300. (a) (5 points) Half way into the month, your smartphone bill is at $100. What is the probability that your bill will be at most $133? U 133) = 133100 = 0.7333 P(U 133|U 100) = P(100 P(U 100) 145100 (b) (5 points) Assume the monthly bills are independent. What is the probability that your monthly bill will exceed $125 at least two times in 24 months. K is number of times bill exceeds $125 in 24 months. K is Binomial(n = 24, = P(U > 125)). Where = 145125 = 0.3333. 14585 P(K 2) = 1 (P(K = 0) + P(K = 1)) = 1 (0.0001 + 0.0007) = 0.9992 (c) (5 points) Assume the monthly bills are independent and we will observe 48 months. What is the probability that your average monthly bill will exceed $118? q U is N ( = 115, = 300 = 2.5) P(U > 118) = P(Z > 48 118115 ) 2.5 = P(Z > 1.2) = 0.1151 (d) (5 points) Assume the monthly bills are independent and we will observe 48 months. What is the probability that the proportion of monthly bills that exceed $125 is greater than 0.35? q P is N (0.3333, 0.33330.6667 = 0.0680) 48 P(P > 0.33) = P(Z > 0.330.3333 0.0680 = P(Z > 0.0485) = 0.5193 3. (20 points) After the Blackfish documentary, SeaWorld decided to make all orca water tanks much larger. Orcas swim about 100 miles per day in the wild, it would take an orca 1,900 laps in a typical tank to reach that length. The argument here is that a larger the pool of water will decrease the amount aggressive behavior. For each whale, SeaWorld recorded aggressive behavior (number of 'aggressive' acts) and the size of current tank (in cubic meters) over a one month span. With the data collected they fit a regression line: y = 805 0.3x (a) (4 points) What do y and x represent in the regression line? y is the estimated number of aggressive attacks. x is the size (in cubic meters) of the orca water tank. (b) (4 points) Interpret the y-intercept and slope of the regression line within the context of the study. y-int: For a water tank of size 0 cubic meters, the estimated number of aggressive acts per month would be 805. (Doesn't apply here). slope. For every one cubic meter increase in the size of water tank, the expected number of aggressive should decrease by 0.3. (Note: could say for every 3 cubic meters we have approx a drop of one aggressive act.) (c) (4 points) Is the correlation coefficient r a positive or negative value? Explain. Negative. The slope of the regression line is negative indicating that as tank size increases, we expect the agressive behavior to decrease. This is a negative linear association. Also b = r ssxy , both sx and sy are always positive, so r must be negative. (d) (4 points) Based from the analysis, SeaWorld claims that the whales will no longer be aggressive due to the larger tanks. Is this a correct explanation? If not, give the correct explanations. No. Correlation does not imply causation. There are lurking variables that could be the actual cause. For example, a park with a larger water tank may have much more money to provide better care for Orca whales. (e) (4 points) An orca activist thought this would be a great opportunity to use SeaWorld's data analysis against them. They want to show that the Antarctic Ocean provides the largest orca tank one can find. All they need to do is estimate the size of the Antarctic Ocean. Is this making good use of the regression line? Can we trust the results? No. We cannot trust the results. This a problem of extrapolation. It is safe to assume that none of the water tanks in the study measure close to the size of the Antarctic. We may only use our regression line within the range of data it was built from. 4. (20 points) The composition of the earth's atmosphere may have changed over time. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen: 63.4 65.0 64.4 63.3 54.8 64.5 60.8 49.1 51.0 Page 2 (a) (2 points) Calculate the sample mean for this data. x = 59.5889 (b) (2 points) Calculate the sample variance for this data. S 2 = 39.1286 (c) (4 points) Compute the 5 number summary for this data. Sort the data: 49.1 51.0 54.8 60.8 63.3 63.4 64.4 64.5 65.0 = 52.9, median = 63.3, Q3 = 64.4+64.5 = 64.45, max = min = 49.1, Q1 = 51+54.8 2 2 65. (d) (6 points) Construct a 95% confidence interval for the mean percent of nitrogen in the atmosphere during the Cretaceous era. Interpret this confidence interval. What assumptions must be met in order for your confidence interval to be valid? We need to assume that percent of nitrogen follows a Normal distribution. 95% CI for mean percentage of nitrogen: ( x tn1,1/2 sn ) = (59.5889 t8,0.975 (59.5889 4.8082) = (54.7807, 64.3971) 39.1286 ) 9 = (59.5889 2.306 39.1286 ) 9 = We are 95% confident that the true mean percentage of nitrogen levels is contained by the interval (54.7807, 64.3971). (e) (6 points) Suppose we want to test if the mean percentage of nitrogen was 62 percent. Perform the hypothesis test with = 0.05. Be sure to list hypotheses, test statistic, rejection region, and conclusion. H0 : = 62 vs Ha : 6= 62 where is the mean percentage of nitrogen. ttest = 59.588962 39.1286 3 = 1.1564 Reject H0 if |ttest | > t8,0.975 . 1.1564 < 2.306, so fail to reject H0 . At a 5% level of significance, there is not enough evidence to reject the claim that the mean nitrogen percentage was 62 percent. We may conclude that mean levels were 62 percent. 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