In article , Terry Harper
writes
There are 13824 three-letter groups available. 3176 (23%) contain an X
or a Z. That roughly matches my perception of their frequency; have you
done a census?
This puzzled me at first, until I realised that there is a 1 in 24 chance of
getting a "Z" in each of one of the columns, which means a 1 in 8 chance of
getting a "Z" in one of the three columns.

No, that's not how you do it. In fact, there's about a 1 in 8.34 chance
of getting a "Z" in one of the three columns, because the cases with two
or three "Z"s pull the odds down.

The way to do it is like this. In each column, there are 22 ways not to
have an X or a Z. So there are 22*22*22 (10648) plates with no X or Z.
There are 24*24*24 (13824) in total, leaving 3176 with an X or Z.

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Clive D.W. Feather, writing for himself | Home:
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In article ,
Colin Rosenstiel writes
I did a rather larger census while cycling from Putney to Westminster this
morning. I saw 154 cars with series 3 index plates; 50 had X or Z (32.5%).

Okay, we can now do some statistical analysis. A census like this is a
binomial distribution B(C,p) where C is the number of cars (you saw 154)
and p is the probability of an X or Z.

I claim that p = 3176/13824 (this is the null hypothesis). You claim it
is larger (the alternative hypothesis).

We approximate B(C,p) by the normal distribution N(Cp, Cp(1-p)), which
is N(0.23N, 0.177N). With X hits (plates with an X or Z), let:

Z = (X - mu)/sigma = (X - 0.23N)/sqrt(0.177N)

and the significance tests say that in a random sample:

Z 1.64 5% of the time
Z 2.33 1% of the time

You had N = 154, X = 50. So Z = (50 - 35.4)/5.22 = 2.79. That's highly
significant (it happens less than 1 in 378 times). It looks like you're
right.

--
Clive D.W. Feather, writing for himself | Home:
Tel: +44 20 8495 6138 (work) | Web: http://www.davros.org
Fax: +44 870 051 9937 | Work:
Written on my laptop; please observe the Reply-To address