Advanced High-School Mathematics

364 CHAPTER 6 Inferential Statistics

1 −(1−p)k = P(Y ≤k)

= P(g(X)≤k)

= P(X≤g−^1 (k))

=

g−^1 (k)

n

Solving forgwe get

g(h) =

ln(1−hn)

ln(1−p)

.

However, we immediately see that ifh =n= nmax we see that

g(h) is undefined (and the TI will generate an error whenever

h = n). This makes mathematical sense as there is no value of

k for which P(Y ≤k) = 1. One remedy is to let the value ofn

in the above expression for g be slightly larger than nmax. Of

course, having done this, the transformed values will no longer be

integers, so we’ll need to round to the nearest integer. On the TI

calculator the value int(x+.5) will have the effect of rounding to

the nearest integer.

Now do this: Generate 100 samples of the geometric random

variable with parameterp=.25 using the command

ln

Å

1 −randInt(1 30. 01 ,^30 ,100)

ã

ln(1−.25)

→L 1

followed by the command

int(L 1 +.5)→L 1.

This will store 100 randomly-generated integer samples in the list

variable L 1. You should check to see if they appear to follow

the geometric distribution with parameter p = .25. (Start by

comparing the mean of your data with the theoretical mean of the

geometric random variable!)