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# Please Help! 50 points!!! (Dont spam my question ill report you)The probability of success in each of the 58 identical engine tests is p=0.92. What is the mean of this binomial distribution?4.353.42.163.0The minimum usual value of a binomial distribution is 89.3. The maximum usual value is 102.8. Which of the following values would be considered usual for this distribution?1038991125A local poll found that only 17% of the 546 people surveyed feel the mayor's new city tax plan is effective. Find the mean of this binomial distribution.77.0453.28.892.836% of the 1035 people who watched the new sequel to a popular action movie thought it was better than the orginal. Find the mean.372.6238.5662.415.4A bonus activity in a popular video game allows the gamer to choose from several boxes to acquire a rare and useful item. The probability of success is 0.2. With 35 attempts, find the standard deviation.2.47.05.60.2The probability of a child being born with light-colored eyes to two particular parents is 0.25. The parents intend to have 5 children. Find the variance of this binomial distribution.0.91.01.33.8The probability of students choosing the left lunch line at school is p=0.431. Find the variance of the binomial distribution of 982 students choosing the left lunch line today.423.2558.815.5240.8A couple of college students, frustrated with the current class registration process, decide to survey 500 random students on campus. They find that 84% are also dissatisfied. What is the standard deviation of this binomial distribution?420.08.267.20.1

Accepted Solution

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Answer:Step-by-step explanation:1)The probability of success in each of the 58 identical engine tests is p=0.92n = 58mean, u = np = 58×0.92 = 53.362) The only value that would be considered usual for this distribution is 91. This is because it is the only value between the minimum and maximum value3) n = 546p = 17/100 = 0.17Mean = np = 546×0.17= 92.824) n = 1035p = 36/100 = 0.36np = 1035 × 0.36 = 372.65) The probability of success is 0.2. p = 0.2q= 1-p = 1-0.2 = 0.8n = 35standard deviation = √npq = √35×0.2×0.8 = 2.346) p = 0.25q = 1-0.25 = 0.75n = 5Variance = npq = 5×0.25×0.75 = 0.97) n = 982p = 0.431q = 1 - p = 1 - 0.431 = 0.569Variance = npq = 982×0.431×0.569= 240.88) n = 500p = 84/100 = 0.84q = 1-0.84 = 0.16Standard deviation = √npqStandard deviation = √500×0.84×0.16 = 8.2