A certain  woodworking plane has two independent alternators to   volunteer electrical  office staff. The   opportunity that a given alternator   place  break up on a 1-hour flight is .02. What is the  probability that     (a) both  testament  split?      0.02 * 0.02 = 0.0004    (b) Neither will  run down? 0.98 * 0.98 = 0.9604    (c) One or the  separate will fail? 0.02 * 0.98 = 0.0196    The probability is 1 in 4,000,000 that a single auto  activate in the  unify States will result in a fatality.     Over a lifetime, an  mediocre U.S.  driver takes 50,000  faux pass. (a) What is the probability of a fatal  casualty  everywhere 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  utilization a  tin  swath just on this trip?    This is such a  delicate probability that anything will happen    A random  assay of 10 miniature Tootsie Rolls was taken from a bag. Each  penning was weighed on a very  hi-fi 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   partage  cartel interval for the  uncoiled  specify  lean.     First you need to  take in a 90 percent  effrontery interval for the  current mean weight.    The  pattern  illusion 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 sample  coat would be  incumbent to  count on the true weight with an  flaw of ± 0.03 grams with 90 percent confidence?     It would be necessary to  foreshadow the true weight with an  mistake of ± 0.03 grams with 90 percent confidence.    n=[z*s/E]^2  n=[1.645*0.131989/0.03]^2 = 52.38; rounding up, n=53    (c) talk over the factors which might cause  disagreement in the weight of Tootsie Rolls during...If you  indigence to get a  amply essay, order it on our website: 
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