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A certain airplane has two independent alternators to bid electrical office staff. The opportunity that a given alternator will break up on a 1-hour flight is .02. What is the luck that (a) both will denounce? 0.02 * 0.02 = 0.0004 (b) Neither will fail? 0.98 * 0.98 = 0.9604 (c) One or the other will fail? 0.02 * 0.98 = 0.0196 The probability is 1 in 4,000,000 that a single auto escape in the unify States will result in a fatality. Over a lifetime, an average U.S. driver takes 50,000 stumblers. (a) What is the probability of a fatal casualty over a lifetime? P= 1-P = 0.00000025 =1-0.00000025 or 0.99999975 0.99999975 to the 50000th power: and 0.9999997550000 = 0.987577799 = 1 - 0.987577799 =0.012422201 Or about 0.0124 (b) why might a driver be tempted not to practice session a tin belt just on this trip? This is such a small probability that anything will demote A random assay of 10 miniature Tootsie Rolls was interpreted fr om a bag. Each penning was weighed on a very(prenominal) accurate scale. The results in grams were 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.

477 (a) generate a 90 role confidence musical interval for the uncoiled mean lean. First you hire to construct a 90 pct confidence interval for the true mean pack. The pattern error is E = 1.96(s/sqrt(n)) = 1.96[0.131989/sqrt(10)]=1.96*0.41739 =0.081808 C.I. = (x-bar-E,x-bar+E) = (3.3048-0.0818,3.3048+0.0818) (b) What render size would be incumbent to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? It would be inevitable to estimate the true weight with an error of ± 0.03 grams with ! 90 percent confidence. n=[z*s/E]^2 n=[1.645*0.131989/0.03]^2 = 52.38; rounding error up, n=53 (c) Discuss the factors which might cause sportsman in the weight of Tootsie Rolls during...If you want to get a luxuriant essay, order it on our website: OrderEssay.net

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