1. No category

 www.LionTutors.com STAT 200 (Kankam) – Exam 2 – Practice Exam Solutions 1. C – No independent because P(CHEM110) times P(MATH140) does not equal P(CHEM110 and MATH140). Independence rule = P(A and B) = P(A) * P(B) P(CHEM110) = 0.4 P(MATH140) = 0.6 P(CHEM110 and MATH140) = 0.28 0.28 ≠ (0.4)(0.6) 2. A – Mutually exclusive The events are mutually exclusive because it is not possible for them both to happen at the same time. If two events are mutually exclusive, you know P(O and B) = 0. The independence rule says that if the two events are independent the following will hold true. P(O and B) = P(O) * P(B) P(0) = 4/6 = 2/3 P(B) = 2/6 = 1/3 0 ≠ (2/3)(1/3) Since the independence rule does not hold, we know the events are not independent. 3. B – No, it is not binomial because the number of trials, n, is not fixed. 4. D – 0.95 P(A or B) = P(A) + P(B) – P(A and B) P(A or B) = 80/100 + 15/100 – 0 = 95/100 5. C – 0.8413 Z = (90 – 60) / 30 = 1 P(Z < 1) = 0. 8413 6. B – Relative frequency probability 7. A – Yes, it is binomial because all of the conditions for a binomial experiment are met. 8. D – 0.85 P(X ≥ 1) = 0.25 + 0.45 + 0.10 + 0.05 = 0.85 9. C – Negative 5/2 Z = (50 – 70) / 8 = –20/8 = –5/2 10. E – Find the probability of 6 success for a binomial variable with n = 10 and p = 1/5. 11. A – Subjective probability 12. D – 0.9332 Z = ($150 – $165) / $10 = –1.5 P(Z > –1.5) = 0.9332 13. C – 13/52 * 13/51 because the probability the first card is a diamond is 13/52. If the first card is a diamond and it is not replaced, there will only 51 cards let in the deck; however, there will still be 13 clubs left in the deck so the second probability is 13/51. 14. C – Orange, orange, orange because orange is the most likely event to occur each time the die is rolled. 15. D – 16/16 because there is a 100% change X will be less than or equal to 4. 16. A – 1/3 because you know the number is one of three possible odd numbers on the die. 17. D – 0.9772 Z = ($150 – $165) / $10 = 2 P(Z < 2) = 0.9772 18. C – Find the cumulative probability for 6 successes for a binomial variable with n = 10 and p = 1/4. 19. A – 0.1587 Z = (90 – 60) / 30 = 1 P(Z > 1) = 0.1587 20. C – 80 E(X) = np = (100)(0.80) = 80 21. E – 5/2 Z = (90 – 70) / 8 = 20/8 = 5/2 22. D – 4/52 * 3/51. There are a total of four 7’s in the deck. The probability the first card is a 7 is 4/52. If the first card is a 7, there are now only three 7’s left in the deck and 51 total cards left in the deck so the second probability is 3/51. 23. D – No, the number of trials is not fixed 24. C – 10 E(X) = (0.5)(20) = 10 25. A – 0.4772 Z = (74 – 68) / 3 = 2 P(Z > 2) = 0.0228 We know that half of the observations will fall below 68 inches and half will fall above 68 inches because 68 inches is the mean. So the area to the right of 68 inches is 0.5000. 0.5000 – 0.0228 = 0.4772 26. D – 12/16 P(X ≤ 2) = 2/16 + 4/16 + 6/16 = 12/16 27. C – 0.5*0.5*0.5*0.5 because the probability the coin lands on tails each time is 50%. 28. B – 60 Look up 0.1587 in table = –1 Cassidy’s score = 70 + (–1)(10) = 60 29. B – No, if two events are independent, knowing that event A occurred will not affect the probability that event B occurs. 30. D – (200)(0.80)(1– 0.80) p = 160 / 200 = 0.80 σ = 𝑛𝑝(1 – 𝑝) =
(200)(0.80)(1 − 0.80) = 5.66 31. D – 0.85 P(X ≤ 2) = 0.15 + 0.25 + 0.45 = 0.85 32. D – 90 Look up 0.9772 in table = 2 Cassidy’s score = 70 + (2)(10) = 90 33. C – 3/4 P(A or B) = P(A) + P(B) – P(A and B) P(A or B) = 3/4 + 1/4 – 1/4 = 3/4 34. A – 0.6915 x = 75 μ = 70 σ = 10 z = (75 – 70) / 10 = 0.5 Look up 0.50 in standard normal table = 0.6915 The problem asked for the proportion of students who scored LOWER than Cassidy so we a looing for P(Z < 0.50), which is 0.6915. 35. A – 1/16 36. B – 0.3085 x = 75 μ = 70 σ = 10 z = (75 – 70) / 10 = 0.5 Look up 0.50 in standard normal table = 0.6915 The problem asked for the proportion of students who scored HIGHER than Cassidy so we a looing for P(Z > 0.50), which is 1 – 0.6915. 37. D – 4/4 P(A or B) = P(A) + P(B) – P(A and B) P(A or B) = 3/4 + 2/4 – 1/4 = 4/4 38. A – 40/65 Total male = 65 Total male in state = 40 Probability = 40 / 65 39. B – 12/20 Probability of not getting selected the first time is 4/5. The probability of not getting selected the second time is 3/4 because one person was already selected (4/5) * (3/4) = 12/20 40. B – Approximately 7% of students spent more than $75 on Jimmy Johns 𝑧=
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 − 𝑀𝑒𝑎𝑛 $75 − $60
=
= 1.5 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
$10
Look up z-­‐score of 1.5 in z-­‐table = 0.9332 0.9332 is the probability of getting a z-­‐score ≤ 1.5; however, we want to find the probability of getting a z-­‐score ≥ 1.5 so we need to subtract 0.9332 from 1. Probability spending more than $75 = 1 – 0.9332 = 0.0668 41. D – 0.95 P(X ≤ 3) = 0.15 + 0.25 + 0.45 + 0.10 = 0.95 42. B – Use the z-­‐table to find the probability of getting a z-­‐score greater than negative 2. 𝑧=
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 − 𝑀𝑒𝑎𝑛 $90 − $120
=
= −2 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
$15
43. B – 0.0668 x = 55 μ = 70 σ = 10 z = (55 – 70) / 10 = –1.5 Our standard normal table only includes positive z values so we will need to look up 1.50 in standard normal table = 0.9332 The value we found in the standard normal table, 0.9332, is the P(Z < 1.50). However, we want to find P(Z < –1.50). Since the normal distribution is perfectly symmetric, we know the area to the right of 1.50 is the same thing as the area to the left of –1.50. P(Z < –1.50) = 0.0668 44. A – 0.9332 x = 55 μ = 70 σ = 10 z = (55 – 70) / 10 = –1.5 Our standard normal table only includes positive z values so we will need to look up 1.50 in standard normal table = 0.9332 The value we found in the standard normal table, 0.9332, is the P(Z < 1.50). However, we want to find P(Z > –1.50). Since the normal distribution is perfectly symmetric, we know the area to the left of 1.50 is the same thing as the area to the right of –1.50. P(Z > –1.50) = 1 – 0.0668 = 0.9332 45. B – Single blinds because the doctor did not know which method each participant used; however, the participants do know which method they used. 46. D – Confounding variable 47. C – 0.3 P(BC) = 1 – P(B) P(BC) = 1 – 0.7 = 0.3 48. C – 0.6826 P(Z ≤ 1) = 0.8413 P(Z ≤ –1) = 0.1587 P(–1 ≤ Z ≤ 1) = 0.8413 – 0.1587 = 0.6826 49. B – 0.0139 P(Z ≥ 2.20) = 1 – 0.9861 = 0.0139 50. D – 0.8849 P(Z > –1.2) = P(Z < 1.2) = 0.8849