exam1 - STAT 350 – Spring 2009 Exam 1 SOLUTION Your Name:

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Unformatted text preview: STAT 350 – Spring 2009 Exam 1 SOLUTION Your Name: ____________________________________ Your Seat: __________ Section Time (circle): 10:30 11:30 4:30 Note: • You are responsible for upholding the Honor Code of Purdue University. This includes protecting your work from other students. • Show your work on all questions. Unsupported work will not receive full credit. Credit will not be given for dumb luck. Showing work includes defining any random variable or event you use in the solution. Showing work also includes identifying any named distribution you are using in the solution and the values of all relevant parameters. • Make it clear what your final answer is to each question. It may help to circle your answer. • Decimal answers should be exact or to at least four significant digits (exceptions as discussed in class). • Standard Normal (Z) values/probabilities must be taken from the table provided. • You are allowed the following aids: a one-page (8.5×11 inch) cheat sheet, a scientific calculator, and pencils. • Turn off and put away your cell phone before the exam begins! Question Points Possible Points Missed 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total 100 Exam Score: _______________/ 100 1. (10 points; 5 points each part) Use the following set of number to answer the questions below. Assume you will be making a histogram of this data using the following cut-points: 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100. 3.695 26.671 43.833 52.710 59.652 70.249 78.219 82.719 87.207 92.490 14.007 28.106 43.981 55.004 61.486 70.682 78.477 83.089 88.380 93.680 17.703 28.805 44.943 55.620 63.623 71.429 78.900 83.852 89.003 93.881 18.947 31.591 45.430 55.669 65.098 72.494 79.272 84.843 89.890 95.002 20.453 33.202 48.031 55.723 65.722 72.660 80.005 85.069 90.252 95.864 20.692 33.939 49.455 56.312 67.247 73.517 81.201 85.271 90.469 97.761 20.723 35.026 49.755 56.748 67.647 75.063 81.700 86.262 91.064 97.811 21.507 39.994 52.470 58.890 70.169 76.971 82.501 87.021 91.169 97.952 a. If you are making a frequency histogram, what would be the height of the bar for the 30 to 40 bin? 5 There are 5 observations (highlighted) with values between 30 and 40 b. If you are making a density histogram, what would be the height of the bar for the 30 to 40 bin? Relative Frequency = (# of observation in that class) / (total # of observations) Density = (relative frequency) / (class width) = (5/80)/10 = 0.00625 2. (10 points; 5 points each part) Base your answers to the questions below on the following data set: 21 33 38 44 47 54 54 55 60 61 66 67 68 69 70 71 72 76 93 a. Find the median of this data set. 61 b. Find the lower quartile of this data set. The lower quartile is the median of the smallest 10 observations (highlighted) LQ is halfway between 47 and 54: (47 + 54) / 2 = 50.5 STAT 350 Exam 1 Page 2 of 2 3. (10 points; 5 points each part) You have a sample of 35 observations from a population, you 35 find that 35 ∑ x = 2288 and ∑ x i =1 i i =1 2 i = 174,526. a. Find x 2288 / 35 = 65.37143 b. Find sx 2 sx = SS xx = n −1 ( 2288) 174526 − 35 − 1 35 2 = 24956.17143 = 734.005042 34 2 sx = sx = 27.09252742 4. (10 points) X is a random variable with the following PDF. Find the mean of X. ⎧0.34 − 0.04 x if 1 < x < 6 f ( x) = ⎨ 0 otherwise ⎩ μ= +∞ ∫ xf ( x ) dx −∞ STAT 350 Exam 1 6 = ∫ x ( 0.34 − 0.04 x ) dx = 37/12 = 3.08333 1 Page 3 of 3 5. (10 points) For Spring Break, 48% of Purdue undergraduates will go to Florida, 22% will go to Mexico and the rest will stay in Indiana. 40% of the undergraduates that go to Florida will get sunburns, 30% of the undergraduates that go to Mexico will get sunburns, and only 1% of the undergraduates that stay in Indiana will get sunburns. What percent of all Purdue undergraduates will get sunburns over Spring Break? P(Burn) = P(Florida)P(Burn | Florida) + P(Mexico)P(Burn | Mexico) + P(Indiana)P(Burn| Indiana) = (0.48)(0.40) + (0.22)(0.30) + (0.30)(0.01) = 0.261 26.1% 6. (10 points) You randomly select 27 tulip bulbs from a large discount bin at your local nursery. The nursery manager tells you that 84% of the bulbs in the bin are for red tulips. If you plants these bulbs and all 27 bloom, what is the probability you will get at least 25 red tulips? X = # of red tulips X ~ Binomial (n = 27, pi = 0.84) n(1 – pi) < 5, therefore the normal approximation to the binomial is NOT appropriate P(X ≥ 25) = P(X = 25) + P(X = 26) + P(X = 27) ⎛ 27 ⎞ 25 2 ⎛ 27 ⎞ 26 1 ⎛ 27 ⎞ 27 0 ⎜ ⎟ .84 .16 + ⎜ ⎟ .84 .16 + ⎜ ⎟ .84 .16 ⎝ 25 ⎠ ⎝ 26 ⎠ ⎝ 27 ⎠ = 0.1704066 STAT 350 Exam 1 Page 4 of 4 7. (10 points; 5 points each part) In the Hoosier Lottery, if you picked 2 of the 6 winning numbers, you win a free “quick pick”. The probability of winning a free quick pick is 0.137. Assume you play the Hoosier Lottery once a week for the next 15 weeks. a. What is the expected value of the proportion of weeks in which you win a free quick pick? E(p) = π = 0.137 b. What is the standard deviation of the proportion of weeks in which you win a free quick pick? SD ( p ) = π (1 − π ) n = 0.137 (1 − 0.137 ) = 0.088781 15 8. (10 points) The lifetime of a particular type of battery has a mean of 19.4 hours with a standard deviation of 4.1 hours. If you have 50 of these batteries, what is the probability that the average lifetime of those 50 batteries is between 18 and 19 hours? E(X1) = 19.4 SD(X1) = 4.1 n = 50 4.1 ⎞ ⎛ X ~ Normal ⎜ μ = 19.4, σ = ⎟ 50 ⎠ ⎝ ⎛ 18 − 19.4 19 − 19.4 ⎞ P (18 < X < 19 ) = P ⎜ <Z < ⎟ = P(−2.41 < Z < −0.69) 4.1 50 ⎠ ⎝ 4.1 50 = Φ ( −0.69 ) − Φ ( −2.41) = 0.2451 − 0.0080 = 0.2371 STAT 350 Exam 1 Page 5 of 5 9. (10 points) Body-mass-index (BMI) is a ratio of weight to height or length. In humans, high values of BMI indicate a person is over-weight (too heavy for their height). In wildlife, BMI (usually measured as a ratio of weight to length) is used as a measure of “condition” individuals with high BMIs are considered healthy and those with low BMIs are considered to be in underweight and, therefore, in poor health (obesity doesn’t happen in the wild). 35 black bears from the George Washington and Jefferson National Forest in Virginia were captured and the BMI was calculated for each bear. For the 35 bears sampled, the average BMI was 246 and the standard deviation was 41. Obtain an 80% confidence interval for the true mean BMI of the population of black bears living in the George Washington and Jefferson National Forest in Virginia. α = 1 – 0.80 = 0.20 zα/2 = Φ-1(1 – 0.20/2) = Φ-1(0.90) = 1.28 80% CI: 246 ± 1.28 41 → 35 (237.12926, 254.8707) 10. (10 points) You wish to estimate the proportion of black bears in the National Forests that are underweight (have a BMI below a particular value). How many bears will you need to sample to ensure that your resulting 95% CI will be within 0.05 of the true value? α = 1 – 0.95 = 0.05 zα/2 = Φ-1(1 – 0.05/2) = Φ-1(0.975) = 1.96 ⎡z ⎤ n = π (1 − π ) ⎢ α /2 ⎥ ⎣B⎦ 2 ⎡ 1.96 ⎤ = 0.5 (1 − 0.5 ) ⎢ ⎣ 0.05 ⎥ ⎦ 2 = 384.16 2 ⎡z ⎤ 384 or 385 are also a π (1 − π ) ⎢ α /2 ⎥ acceptable answers ⎣B⎦ STAT 350 Exam 1 Page 6 of 6 ...
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This note was uploaded on 02/16/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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