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Normal Distribution


Date: 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
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Associated Topics:
High School Probability
High School Statistics

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