(1)

(2)

(3)

(4)

(5)

0.8333

0.1672

3

5!/(3!2!)=10.0

0.159

where

0.8334

0.1671

4

5!/(4!1!)=5.0

0.396

0.410

5

0

5

5!/(5!0!)=1.0

0.833

0.167

0.96

*

Binomial coefficient.

**

Product of columns 2,3,4.

Given that the 0.833 nonexceedance probability event

Consequently, there is a 0.96 probability, or 96 percent

or 0.167 exceedance probability event occurs, the

chance, that the stage corresponding to the 0.167-

the third-ranked observation, *Y*j = Y3 can be computed

event *Y*3.

using Equation 1:

DISTRIBUTION DUE

TO APPLICATION OF

VARIABLE

INCOMPLETE BETA

FUNCTION

Xn

PREDICTION

LIMITS

X

1

1.0

0.95

O.5

0.05

0.0

EXCEEDANCE PROBABILITY

3