Express the confidence interval −29.8<μ<265.6-29.8<μ<265.6 in the form of ¯x±MEx¯±ME. ¯x±ME=x¯±ME=
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Express the confidence interval −29.8<μ<265.6-29.8<μ<265.6 in the form of ¯x±MEx¯±ME.
¯x±ME=x¯±ME=
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Once it has failed, the mean time to repair is 2.5 days (the repair rate is constant). Calculate the interval availability over a 2-day mission (starting at time zero) and the point availability at the end of the 2 days.Assume X is normally distributed with a mean of 13 and a standard deviation of 2. Determine the value for x that solves each of the following. Round the answers to 2 decimal places. a) P(X> x) = 0.5. x = b) P(X > x) = 0.95. X = c) P(x < X < 13) = 0.2. X = i d) P(-x < X 13 < x) = 0.95. X = i e) P(-x < X - 13 < x) = 0.99. x = iIf X is normally distributed and P(X<20)=0.5 and P(X>26)=0.0228, find the mean and standard deviation of X. Give a detailed solutionA sample of 55 measurements resulted in a sample mean, ?⎯⎯⎯=9.3x¯=9.3 and sample standard deviation ?=1.47s=1.47. Using ?=0.05α=0.05, test the null hypothesis that the mean of the population is 10.510.5 or less against the alternative hypothesis that the mean of the population, ?>10.5μ>10.5 by giving the following:(a) the degree of freedom (b) the test statistic The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that ?=10.5μ=10.5. B. We can reject the null hypothesis that ?=10.5μ=10.5A herbal extract is claimed to lower the blood glucose level. 20 prediabetic subjects were give capsules containing lactose for 1 month and their mean blood glucose level (BGL) was measured as 127±10 mg/dl. Another sample of 20 subjects were given capsules containing dried herbal extract for one month was and the mean BGL was measured as 121±12. The question of interest was does the herbal extract result in significant BGL reduction compared to placebo. ASsuming the populations have equal standard deviation, test statistic is calculated as O 4.318 O 5.412 O 4.654 O 3.123 O 2.745 O 1.718 O 1.34 O 2.214Suppose that the heart beat per minute (bpm) of adult males has a normal distribution with a mean of mu equals 73.4μ=73.4 bpm and a standard deviation of sigma equals 11.6σ=11.6 bpm. Instead of using 0.05 for identifying significant values, use the criteria that a value x is significantly high if P(x or greater)less than or equals≤0.01 and a value is significantly low if P(x or less)less than or equals≤0.01. Find the pulse rates for males that separate significant pulse rates from those that are not significant. Using these criteria, is a male pulse rate of 90 beats per minute significantly high?Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. 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