At the 2018 Winter Olympic Games, figure skater Mirai Nagasu became the first Am

Question

At the 2018 Winter Olympic Games, figure skater Mirai Nagasu became the first American woman to land the triple axel in Olympic competition. The triple axel very sophisticated jump. To prepare for this, she practiced it once every day for a number of a years. After all this practicing, she lands the triple axelwth a probability of 0.87. Assume that each attempt is independent. a) In a training session for the Olympics, Mirai will practice the triple axel until she successfully lands it. What 4. is the probability she lands her first successful jump on her 5th attempt? What distribution, parameter(s), and support are you using? What is probability that it takes Mirai at most 4 attempts to land her first successful triple axel? Mirai has just made her fifth attempt and has not yet been successful in landing it. What is the probability it takes Mirai at least 9 attempts to land her first triple axel? Later in the week her coach demands that Mirai should not attempt the triple axel at the Olympics unless Mirai can land it 10 times successfully. What is the probability Mirai can complete this task in 12 attempts? What distribution, parameter(s), and support are you using? What is the expected value and standard deviation of the number of attempts it will take for Mirai to land her 10th successful triple axel? b) c) d) e)

Explanation / Answer

a) For this situation, a binomial distribution is suitable. The parameters are success probability p = 0.87, failure probability 1 - p = 0.13 and number of trials n = 5.

The probability that she will be successful only in her last (fifth) attempt = 0.13 * 0.13 * 0.13 * 0.13 * 0.87 = 0.0002484807 (round up these answers as necessary)

b) She could do it in one attempt itself. Or in two. Or in three. Or in four. So there are four 'or' cases for which probabilities of each have to be added (cumulative probabilty). P(success in at most 4 attempts) = P(success in 1st attempt) + P(success in 2nd attempt after failing 1st) + P(success in 3rd attempt after failing 1st and 2nd) + P(success in 4th attempt after failing 1st, 2nd and 3rd) = 0.87 + (0.13 * 0.87) + (0.13 * 0.13 * 0.87) + (0.13 * 0.13 * 0.13 * 0.87) = 0.99971439  

c) P (at least 9 attempts with success on last) = 0.000001

d) P(10 successes out of 12 attempts) = 12C10 * (0.87)^10 * (0.13)^2 = 0.27709147619, using binomial distribution.

e) Expected value = 10 * 0.24842341419 + 11 * 0.35524548229 + 12 * 0.27709147619 + 13 * 0.15609486492 + 14 * 0.07102316354 + 15 * 0.02769903378 + 16 * 0.00960233171 + (and so on)...= 13.31 (approximately).

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