Wikipedia:Reference desk/Archives/Mathematics/2009 May 23

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May 23

A LaTeXnicality

I used \renewcommand to make the "enumerate" environment label each item with a capital letter followed by a period, thus:

\renewcommand{\labelenumi}{\Alph{enumi}{.}}

\begin{enumerate}

\item so there

\item and so on

\item blah blah

\item \label{thisgap}

\end{enumerate}

We will see that any gap of the kind described in (\ref{thisgap}) corresponds to a 

I expected a sentence that says

We will see that any gap of the kind described in (D) corresponds to a

but instead I got this:

We will see that any gap of the kind described in (4) corresponds to a

How can I get it to show a (D) rather than a (4)? Michael Hardy (talk) 02:17, 23 May 2009 (UTC)

This does the job:

\documentclass{article}
\usepackage{enumerate}
\begin{document}
\begin{enumerate}[A.]
\item so there
\item and so on
\item blah blah
\item \label{thisgap}
\end{enumerate}
We will see that any gap of the kind described in (\ref{thisgap}) corresponds to a
\end{document}

 — Pt _(T) 22:01, 23 May 2009 (UTC)

It worked. Thank you. Michael Hardy (talk) 05:02, 25 May 2009 (UTC)

Probability of dots on a circumferance

I've read a popular book that mentions in passing calculating the randomness of dots on a circumferance of a circle. The details given are where seven dots lie within an arc of 24 degrees, and four other dots lie outside that range. 24/360 is 1/15. The formula given for calculating the probability is p = (1/15)^7 (14/15)^4 11!/(7!4!) = 1.5 x 10^-6.

Can someone explain how this formula was obtained please, so that I could use it with different numbers of dots?

The formula was used to test the periodicy of the recurring peaks in a time series, by having the time run repeatedly around the circle rather than along the more usual straight line. Are there any other formula for testing the randomness of peaks in a time series? 78.149.172.201 (talk) 13:02, 23 May 2009 (UTC)

Look up Binomial distribution here on wikipedia or in that text book. $(1/15)^{7}(14/15)^{4}{\frac {11!}{7!4!}}=(1/15)^{7}(1-1/15)^{4}{11 \choose 7}$ —Preceding unsigned comment added by Taemyr (talk • contribs) 13:52, 23 May 2009 (UTC)

This looks like a case of testing a hypothesis suggested by the data. A better statistical test would take into account the precise locations of the 11 points. Michael Hardy (talk) 11:06, 24 May 2009 (UTC)

And what would that better statistical test be, please? 78.147.247.28 (talk) 13:34, 26 May 2009 (UTC)