Normal DistributionDate: 6/24/96 at 7:28:46 From: Anonymous Subject: Normal Distribution H~S(5.5,0.5^2), S~N(5.1,0.6^2) and J~N(6.2,0.8^2), H is the distance covered by the hop of a triple jump, in meters. S is the distance covered by the skip of a triple jump, in meters. J is the distance covered by the jump of a triple jump, in meters. H,J,S are independent. In six successive independent 'triple jumps' what is the probability that at least one total distance will exceed 18 meters? Date: 6/24/96 at 8:5:56 From: Doctor Anthony Subject: Re: Normal Distribution The total T = H + S + J. The mean will be E(H) + E(S) + E(J) = 5.5 + 5.1 + 6.2 = 16.8. Variance of total = 0.5^2 + 0.6^2 + 0.8^2 = 1.25 s.d of total = sqrt(1.25) = 1.118 In one jump the probability that T > 18 is given from Normal tables using z = (18-16.8)/1.118 = 1.0733. The table gives A(z) = 0.8584, and we require the tail area beyond z = 1.0733. This area = 1 - 0.8584 = 0.1416. So for each jump, the probability of exceeding 18 metres is 0.1416. In 6 jumps, the probability of at least one jump over 18 metres is 1 - prob.(no jump over 18 m) 1 - 0.8584^6 = 1 - 0.4000 = 0.6000 The probability of at least one jump over 18 metres in six goes is 0.6000. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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