6:42 PM webassign.net o AT&T; LTE Need Help? Talk to a Tutor -/12 points 6.2016
ID: 3302764 • Letter: 6
Question
6:42 PM webassign.net o AT&T; LTE Need Help? Talk to a Tutor -/12 points 6.2016 My Note Richard has just been given a 10-question multiple-choice quiz in his history class. Each question has five answers, of which only one is correct. Since Richard has not attended class recently, he doesn't know any of the answers Assuming that Richard guesses on all ten questions, find the indicated probabilities. (Round your answers to three decimal places.) (a) What is the probability that he will answer all questions correctly? (b) What is the probability that he will answer all questions incorrectly? (c) What is the probability that he will answer at least one of the questions correctly? Compute this probability two ways. First, use the rule for mutually exclusive events and the probabilities shown in the binomial probability distribution table Then use the fact that P(r 1) = 1-P(r = 0) Compare the two results. Should they be equal? Are they equal? If not, how do you account for the difference? O They should be equal, but differ substantially O They should be equal, but may not be due to table error They should be equal, but may differ slightly due to rounding error O They should not be equal, but are equal (d) What is the probability that Richard will answer at least half the questions correctly? Need Help?-Reade-LaiktoaTutori Submit Answer Save Progress Pracice AnotherVinVewng Saved Work Revart to Last Response /6 points Previous Answers BBBasicStat7 6.3.005. Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean to answer the following questions My Not (a) Would it be unusual to have more than 120 successes out of 200 trials?Explanation / Answer
(According to Chegg policy, only four subquestions will be answered. Please post the remaining in another question)
(a) Probability that Richard answers each question correctly = 1/5
=> Probability that Richard answers all questions correctly = (1/5)10 = 0.210 = 0.0000001024
(b) Probability the Richard answers each question incorrectly = 4/5
=> Probability that Richard answers all questions incorrectly = (4/5)10 = 0.810 = 0.1074
(c) P(r >= 1) = 1 - P(r = 0)
= 1 - 0.1074
= 0.8926
They should be equal but may differ slightly due to rounding error.
(d) P(r = 0) = (4/5)10 = 410 / 510
P(r = 1) = (1/5)(4/5)9 = 49 / 510
P(r = 2) = 48 / 510
and so on
=> P(r >= 5) = 45 / 510 + 44 / 510 + 43 / 510 + 42 / 510 + 4 / 510 + 1 / 510
Using sum of GP formula, 1 + 4 + 42 + 43 + 44 + 45 = (46 - 1) / (4 - 1) = 4095 / 3 = 1365
=> P(r >= 5) = 1365 / 510 = 0.00014
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