University:
Stony Brook University
Course:
AMS 310 | Survey of Probability and Statistics
Academic year:

2024
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469

Pages:

7
Author:

study_genius

AMS 310 Spring 2024 Test 2 Form B Solutions Fred Rispoli There are 10 problems, each is worth 10 points. You must show work for each problem. 1. The time (in minutes) that it takes a car mechanic to change the transmission fluid is a random variable X that follows an exponential distribution with mean 30 minutes. a) Find P(14 < X < 22). Round off your answer to three decimal places. (5 pts) 22 −x −x 1 30 22 e dx =−e 30 ¿14=−e ∫ 30 14 −22 30 +e −14 30 =0. 147 ¿ −22 30 −14 30 P ( x<22 )−P ( x <14 )=(1−e )−( 1−e )=0.147 b) Find the 90th percentile. Round off to the nearest whole number. (5pts) x −t 1 0.9=∫ e 30 dt 0 30 −x 0.9=1−e 30 −x e 30 =0.1 −x =ln ⁡(0.1) 30 x=69.08  69 2. Suppose that f(x,y) is a joint uniform continuous distribution where 0 < x < 3 and 1 < y < 5. a) Find the pdf f(x,y) (3pts) This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:55 GMT -06:00 https://www.coursehero.com/file/230816414/AMS-310-Test-2B-Spring-2024-Solutionsdocx/ f ( x , y )= 1 1 = area 12 b) Find P(x > 1, y < 3). Round off your answer to two decimal places. (3pts) 3 3 3 1 1 1 dydx=∫ dx= =0.33 ∫∫ 12 3 1 1 1 6 c) Determine if X and Y are independent. Show your reasoning. (4pts) f ( x )= 1 3 f ( y )= 1 4 1 ∗1 1 3 f ( x , y )= = =f ( x ) f ( y ) 12 4 X ∧Y areindependent 3. In a certain city, the number of power outages per year is a random variable, having a distribution with mean 12 and standard deviation 3.5. If this distribution is normal, what is the probability that: a) There will be between 13 and 15 outages in any year? Round off to the nearest percent. P ( 13< x< 15 )=P 15−12 1  x = 1 ), then find the exact value. (5pts) 2 2 3 y + )dy=¿ ¿ 6/10 = 0.6 ∫ ( 10 5 1 7. The average length of a swordfish is 10 feet and follows a normal distribution with a standard deviation of 1.75 feet. If X represents the mean length of a random sample of 8 swordfish, find P( X > 11)? (10pts) ( P ( X >11 )=1−P z < 11−10 =1−P ( z <1.62 ) 1.75 = 1-0.9474 = 0.0526 √8 ) 8. a) An investigator is studying times between the call for an ambulance and a patient's arrival at a hospital. A T statistic is used with a sample size of 11. Find P(T < 1.812). (5pts) P ( T <1.812 ) =1−P ¿ This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:55 GMT -06:00 https://www.coursehero.com/file/230816414/AMS-310-Test-2B-Spring-2024-Solutionsdocx/ df =10 b) Find a 95% confidence interval for the time between the call for an ambulance and a patient’s arrival at hospital if a random sample of size 11 has a sample mean of 20 minutes and a sample s standard deviation of 3 minutes. Use the formula x  t* ( ). √n x  t* ( s 3 ¿ = 20  2.015 ) = 20  2.228( √n √11 (17.985, 22.015) 9. At a gas station, the fraction of customers who purchase high octane gas in a week is Beta ¿). (a) If the gas station serves an average of 1,000 customers per week, how many customers are expected to purchase high octane gas next week? Round to the nearest whole number. (b) Find the pdf needed to calculate the probability that more than 25% of the customers will purchase high octane gas over the next week and set up the required integral. Do not evaluate the integral. f ( x )= Γ ( 2+5 ) 2−1 6! x (1−x)5−1= x (1−x )4=30 x (1−x)4 , 0< x <1 4! Γ ( 2) Γ ( 5) 1 4 P ( X >1/4 )= ∫ 30 x (1−4 ) dx 1/4 a) E ( X )= α 2 = =2/7 α + β 2+5 The expected number of such customers is 1000 ( 2/7 )=286 . f ( x )= Γ ( 2+5 ) 2−1 6! x (1−x)5−1= x (1−x )4=30 x (1−x)4 , 0< x <1 4! Γ ( 2) Γ ( 5) 1 4 P ( X >1/4 )= ∫ 30 x (1−4 ) dx 1/4 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:55 GMT -06:00 https://www.coursehero.com/file/230816414/AMS-310-Test-2B-Spring-2024-Solutionsdocx/ 10. Suppose that X has a binomial distribution with n = 150 and the success probability p = 0.4. Find approximations to the following probabilities. a) P(61 ≤ X < 66) b) P(X ≥ 64) First note that μ=np=150 ( 0.4 )=60∧n ( 1− p ) =150 ( 0.6 )=90 is large. So we can use the normal approximation to the binomial. σ =√ np ( 1− p ) =√150 ( 0.4 ) ( 0.6 )= √36=6 (a) P ( 61≤ X <66 ) ≈ P 65.5−60 63.5−60 =P ( Z> 0.58 )=1−Փ ( 0.58 )=1−0.7190 6 ) ¿ 0.281 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:55 GMT -06:00 https://www.coursehero.com/file/230816414/AMS-310-Test-2B-Spring-2024-Solutionsdocx/ Part of the Normal Distribution Table z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.9452 0.9554 0.9641 0.9713 0.9772 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 0.9778 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.9474 0.9573 0.9656 0.9726 0.9783 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485 0.8708 0.8907 0.9082 0.9236 0.9370 0.9484 0.9582 0.9664 0.9732 0.9788 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508 0.8729 0.8925 0.9099 0.9251 0.9382 0.9495 0.9591 0.9671 0.9738 0.9793 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531 0.8749 0.8944 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 0.9744 0.9798 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.8770 0.8962 0.9131 0.9279 0.9406 0.9515 0.9608 0.9686 0.9750 0.9803 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 0.8980 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.9808 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761 0.9812 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 Part of the T-Distribution df 0.25 0.2 10 11 12 0.700 0.697 0.695 0.879 0.876 0.873 Right tail probability 0.15 0.1 1.093 1.088 1.083 1.372 1.363 1.356 0.05 0.025 0.02 0.01 0.005 1.812 1.796 1.782 2.228 2.201 2.179 2.359 2.328 2.303 2.764 2.718 2.681 3.169 3.106 3.055 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:55 GMT -06:00 https://www.coursehero.com/file/230816414/AMS-310-Test-2B-Spring-2024-Solutionsdocx/ Powered by TCPDF (www.tcpdf.org)