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(HP-65) Binomial distribution with large input support
06-25-2022, 08:42 PM (This post was last modified: 06-25-2022 08:59 PM by Thomas Klemm.)
Post: #2
RE: (HP-65) Binomial distribution with large input support
As a rule of thumb the binomial distribution can be approximated by the normal distribution if both \(np > 5\) and \(nq > 5\).
This is given in the example since:

\(
\begin{align}
4000 \cdot \frac{1}{6} &\approx 666.6667 > 5 \\
4000 \cdot \frac{5}{6} &\approx 3333.3333 > 5 \\
\end{align}
\)

We can calculate mean \(\mu\) and standard deviation \(\sigma\) with:

\(
\begin{align}
\mu &= n \cdot p \\
\sigma &= \sqrt{n \cdot p \cdot q} \\
&= \sqrt{n \cdot p \cdot (1 - p)} \\
\end{align}
\)

4000
ENTER
6
1/x
×
STO 0

666.66667

1
LSTx
-
×
\(\sqrt{x}\)

STO 1

23.570226

The following program for the HP-15C can be used to integrate the pdf of the normal distribution:
Code:
   001 {    42 21 11 } f LBL A
   002 {       43 11 } g x²
   003 {           2 } 2
   004 {          10 } ÷
   005 {          16 } CHS
   006 {          12 } eˣ

In order to get the best approximation we subtract 0.5 from \(x\) or add 0.5 to \(x\).

FIX 5

661.5
RCL - 0
RCL ÷ 1

-0.21920

672.5
RCL - 0
RCL ÷ 1

0.24749

\(\int_y^x\) A

0.46244


Now we just have to compensate for the missing factor \(\frac{1}{\sqrt{2 \pi}}\) in the pdf:

2
\(\pi\)
×
\(\sqrt{x}\)
÷

0.18449


I haven't checked but I assume that there are already programs for the HP-65 that deal with the normal distribution.
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RE: (HP-65) Binomial distribution with large input support - Thomas Klemm - 06-25-2022 08:42 PM



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