Grand Canyon University:PSY 520 Topic 2 Exercise,Chapter 5 and 8;Verified Answers(Graded A)

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PSY 520 Topic 2 Review Questions Chapter 5 5.13; IQ scores on the WAIS test approximate a normal curve with a mean of 100 and a standard deviation of 15. What IQ score is identified with: a. The u... pper 2 percent X = Mean + (Z)(Standard Deviation) X = 100 + (2.05)(15) X = 100 + 30.75 X = 130.75 The upper 2% IQ score is 130.75 b. The lower 10 percent X = Mean + (Z)(Standard Deviation) X = 100 + (-1.28)(15) X = 100 + (-19.2) X = 100 – 19.2 X = 80.8 The lower 10% IQ score is 80.8 and below c. The upper 60 percent X = Mean + (Z)(Standard Deviation) X = 100 + (-0.25)(15) X = 100 + (-3.75) X = 100 – 3.75 X = 96.25 d. The middle 95 percent 95% = Mean – 2(Standard Deviation) to Mean + 2(Standard Deviation) 100 – 2(15) to 100 + 2(15) 100 – 30 to 100 + 30 70 to 130 The middle 95% ranges from an IQ score of 70 to 130 e. The middle 99 percent X = Mean + (Z)(Standard Deviation) This study source was downloaded by 100000831988016 from CourseHero.com on 05-06-2022 04:05:14 GMT -05:00 https://www.coursehero.com/file/94940505/PSY-520-Topic-2-Review-Questionsdocx/ X = 100 + (-2.58)(15) X = 100 + (-38.7) X = 100 – 38.7 X = 61.3 X = 100 + (2.58)(15) X = 100 + (38.7) X = 138.7 The middle 99% ranges from an IQ score of 61.3 to 138.7 5.14; For the normal distribution of burning times of electric light bulbs, with a mean equal to 1200 hours and a standard deviation equal to 120 hours, what burning time is identified with: a. Upper 50% If the median = 1200, Upper 50% burning time would be anything 1200 hours and greater. b. Lower 75% X = Mean + (Z)(Standard Deviation) X = 1200 + (0.67)(120) X = 1200 + 80.4 X = 1280.4 Lower 75% burning time is 1280.4 hours or less. c. Lower 1% X = Mean + (Z)(Standard Deviation) X = 1200 + (-2.33)(120) X = 1200 + (-279.6) X = 1200 – 279.6 X = 920.40 Lower 1% burning time is 920.4 hours d. Middle 90% X = Mean - (Z)(Standard Deviation) X = 1200 + (1.64)(120) X = 1200 + 196.8 X = 1003.2 This study source was downloaded by 100000831988016 from CourseHero.com on 05-06-2022 04:05:14 GMT -05:00 https://www.coursehero.com/file/94940505/PSY-520-Topic-2-Review-Questionsdocx/ X = Mean + (Z)(Standard Deviation) X = 1200 + (1.64)(120) X = 1200 + 196.8 X = 1396.8 The middle 90% burn time ranges from 1003.20 hours to 1396.8 hours Chapter 8 8.16; A traditional test for extrasensory perception (ESP) involves a set of playing cards, each of which shows a different symbol (circle, square, cross, star, or wavy lines). If C represents a correct guess and I an incorrect guess, what is the probability of: (work in attached picture) a. C? 1 out of 5 or 20% chance b. CI (in that order) for two guesses? 4 out of 25 or 16% chance c. CCC for three guesses? 1 out of 125 or a o.008% chance (so less than a 1% chance) d. III for three guesses? 64 out of 125 or a 51.2% chance 8.19; A sensor is used to monitor the performance of a nuclear reactor. The sensor accurately reflects the state of the reactor with a probability of .97. But with a probability of .02, it gives a false alarm (by reporting excessive radiation even though the reactor is performing normally), and with a probability of .01, it misses excessive radiation (by failing to report excessive radiation even though the reactor is performing abnormally). Known info: - accurate probability = .97 - false alarm probability = .02 - misses radiation excessively = .01 a. What is the probability that a sensor will give an incorrect report, that is, either a false alarm or a miss? This study source was downloaded by 100000831988016 from CourseHero.com on 05-06-2022 04:05:14 GMT -05:00 https://www.coursehero.com/file/94940505/PSY-520-Topic-2-Review-Questionsdocx/ Probability (P) of a false alarm (.02) or a miss (.01) P = .02 + .01 P = .03 or 3% The probability that the sensor will produce an incorrect report or false alarm is 3% or .03 b. Independent sensor introduced; reactor shut down only when both sensors report excessive radiation. What is the new probability that the reactor will be shut down because of simultaneous false alarms by both the first and second sensors? Probability (P) of simultaneous false alarms by both the first and second sensors (0.02) P = 0.02 x 0.02 P = 0.0004 The probability of a simultaneous false alarm by both the first and second sensors is a . 0004 chance or .04% c. Proposed shut down of reactor whenever either sensor reports excessive radiation. What is the new probability that excessive radiation will be missed simultaneously by both the first and second sensor? Probability (P) of simultaneous missed alarm by both the first and second sensors (0.01) P = 0.01 x 0.01 P = 0.0001 The probability of a simultaneous missed alarm by both the first and second sensors is a . 0001 chance or .01% [Show More]

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