Wikipedia:Reference desk/Archives/Mathematics/2008 October 7

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October 7

Name

What's the name given to numeric 6's and 9's that have their thingos swinging out, like the ones in this font? February 15, 2009 (talk) 00:23, 7 October 2008 (UTC)

What font? Wikipedia text is shown in your browser's default sans-serif font, which is probably not the same as my browser's default sans-serif font. Anyway, perhaps you're referring to text figures? —Ilmari Karonen (talk) 04:23, 7 October 2008 (UTC)

I believe that "seriff" is the "feet" on letters such as i, l, t etc... But I believe the rule might be similar to whether or not a lowercase Q or a lowercase J gets curved at bottom, or just a straight line. Guessing from memory (and yes I had to learn this for a computer class) it might be Times, but there's also a rule of whether or not a W gets the same amount of width, as a lowercase L. Its interesting stuff, but your question about 6's for example, is probably the same answer as to the stipulation about lowercase Q and lowercase J. That might help you find it easier, since you can expand your question, i.e. a google search. Sentriclecub (talk) 10:45, 7 October 2008 (UTC)

Try the article on font. Sentriclecub (talk) 10:46, 7 October 2008 (UTC)

Swash (typography) that should be it. Each typeface has its own rules of whether or not to give swashes onto a 6 or 9. Sentriclecub (talk) 10:56, 7 October 2008 (UTC)

Also see Regional handwriting variation#Arabic numerals --71.106.183.17 (talk) 04:36, 10 October 2008 (UTC)

Compass and Midpoint

I recently encountered a problem related to geometry drawing, but did not find an article on it.

So basically, you have a compass. No ruler or anything else. This means you cannot draw a straight line. So given a line segment, use the compass to find the midpoint of it.

Anyways if there is any relevance in this, I thought it might be important to write an article related to geometry drawing? Any tips on the problem let me know ;).

--electricRush (T C) 03:23, 7 October 2008 (UTC)

Are you absolutely sure that you do not have access to a straightedge? Because if you did, you could draw two congruent circles centered on the endpoints of the line segment. Then you would draw a line between the points of intersection between these circles, and then that line would intersect with the original segment at its midpoint. 206.117.40.10 (talk) 03:55, 7 October 2008 (UTC)

Yes, for this problem you may not use a straightedge. All you can do is use the compass to draw circles. This problem is supposed to be very hard though... --electricRush (T C) 04:08, 7 October 2008 (UTC)

Can you access this file (PS)? It describes how to perform the operation. It also speaks of the (for me unknown until tonight) Mohr-Mascheroni theorem. Pallida  Mors 06:52, 7 October 2008 (UTC)

In case you can't handle the postcript, here there's a web site. Pallida  Mors 07:04, 7 October 2008 (UTC)

Even more amazingly see Weisstein, Eric W. "Matchstick Construction". MathWorld., you only need matchsticks to do anything a normal compass and straightedge can do. Dmcq (talk) 07:55, 7 October 2008 (UTC)

Infinite increase

If I have an integer, let's take 1, and I increase that integer at an infinite speed (or rate), does my number become infinite in a zero time period? —Anonymous Dissident^Talk 07:40, 7 October 2008 (UTC)

It's hard to increase an integer; they're pretty stubborn about staying just what they are. (An integer variable is another matter.)

Extrapolating from your question, it's quite possible for a function to have infinite first derivative at a point and yet be pretty well behaved generally. (OK, technically, the derivative won't be defined at that point, but it's pretty obvious what "infinite first derivative" means.) An example is the cube-root function, whose derivative becomes infinite at zero, but (the original function) takes only finite values on the real line. --Trovatore (talk) 07:54, 7 October 2008 (UTC)

From the concept of a massless string, the answer is yes. If you apply a net force to a massless string, it will accelerate infinitely and reach an infinite velocity, instantly. But you can't apply a force, before the force is applied, which is why your Q is a bit tricky. As soon as the instant you apply your force, say at t=2.7 seconds, then yes it instantly reaches infinite velocity. Oops, upon second thought, you would need an infinite acceleration, not an infinite speed (or rate). But since you're asking about the number reaching an infinite "position" (which i'm butchering your question now), then yes if you apply an infinite velocity to an object, it will instantly reach an infinite position. Just the same that a tiny net-force applied to a massless string will cause it to reach an instant infinite velocity, then the reasoning holds true about your original question. So the answer is yes but you could better clearify the last 5 words, since the word "instantly" is what you are asking, right? Sentriclecub (talk) 10:39, 7 October 2008 (UTC)

Actually the answer is "no" (or more exactly, "not necessarily"). That's what my first answer said, but apparently I didn't say it clearly enough. --Trovatore (talk) 12:50, 7 October 2008 (UTC)

Correlation between shapes

I am looking for a formula which determines a correlation coefficient which measures the correlation between two shapes in a two-dimensional space. Each shape is given by connecting the heads of six vectors all pointing in independent directions with angles of 2*pi/6 radians between them. There are two shapes. If the two shapes are identical then the correlation should be one. If the shapes are pointing in opposite directions we would expect the correlation to be negative. The correlation coefficient should between minus one and plus one. Does anybody know how to construct such a formula? HowiAuckland (talk) 10:34, 7 October 2008 (UTC)

I'm afraid you can't. Think again what 'the shapes are pointing in opposite directions' means.

Imagine a regular hexagon, centered at the coordinate system's origin. Let P and Q be two opposite vertices of the figure. Make two new shapes by moving P and Q, respectively, a bit outside from their original positions. I suspect you would call it 'shapes pointing in opposite directions', thus assigning a correlation of −1 to that pair. However if the vertices displacement is small enough, the shapes stay 'almost' identical to each other (and to the original hexagon) so their 'correlation' ought to be close to +1.

So we have a pair of shapes which deserve a correlation of −1 and +1 at the same time. How are you going to solve it?

CiaPan (talk) 13:28, 7 October 2008 (UTC)

Hi CiaPan. Thanks for your reply and you may very well be correct in your assertion. However, in your example just noted the second polygon isn't actually pointing in an opposite direction as the first polygon. In fact, the second polygon is equally pointing, the same amount, in both opposite directions, which is very similar to the first polygon, so I would expect the correlation to be close to +1. Your example has no assymetry, and therefore I expect the correlation to be close to +1. The negative correlation should come from an asymmetry between the two polygons. An example of two polygons which I think should give a negative asymmetry is if you took your example, and instead of stretching out both vertices P and Q equally, you simply stretch one side, say P. Now I would think this would have a negative correlation between these two shapes.

A quandary that I have come across which has made me think that the problem might not be well-defined is if you take as your first polygon, a polygon with P stretch outward significantly in one direction, and the second polygon stretched outward at Q in the opposite direction, then you should get a negative correlation. However if you then look at the limit as the P and Q vertices are shrunk back to the same length as the other vertices from the center, then the correlation should get progressively negative but closer to zero. The quandary is that in that limit one might also expect the correlation to approach +1 since in that limit the two shapes are identical. Does anybody think that this example also indicates the possible non-well-posedness of the problem? HowiAuckland (talk) 20:53, 7 October 2008 (UTC)

You could calculate the mean of the six points and measure the angle between the means for each shape (considered in polar co-ordinates). That would give you a number between 0 and pi, call it x. Then consider 1-2x/pi, does that satisfy your requirements? It would consider two shapes that are just simple scalings of eachother to be identical, is that a problem? --Tango (talk) 23:09, 7 October 2008 (UTC)

This quantity would be undefined when the means for either shape add up to zero. HowiAuckland (talk) 23:36, 7 October 2008 (UTC)

That's a good point, but I think that's due to your question. You talk about whether or not they are pointing in the same direction, but a regular hexagon doesn't point in any direction. --Tango (talk) 23:49, 7 October 2008 (UTC)

Very true. Regular hexagon wouldn't point in any direction. However, the correlation between two identical shapes should always be +1, and if they are very similar it should be close to +1. (I think, subject to the quandary I listed above, which I can't resolve.) HowiAuckland (talk) 23:58, 7 October 2008 (UTC)

If you're willing to skip regular hexagons, you could just literally use the correlation of the two sextuples of vector lengths. Then any other shape would have correlation 1 with itself and would typically have negative correlation with itself rotated or reflected. But you're not going to get perfect -1 in general. --Tardis (talk) 00:37, 8 October 2008 (UTC)

Yes, I think this might be a good idea, except it seems to give strange results. As an example, in the case where you have all the lengths zero except one vector equal to 1 in the first shape, and then in the second shape you have all lengths to be zero except for the opposite direction with length of 1. Then this gives a correlation of −0.2. Similarly, if you take all lengths to be zero except for a single entry, regardless of position, to be one, you will always get a correlation of −0.2, even when the vectors in the two shapes are pointing nearly in the same direction. I would expect this case to give a positive correlation. Even though the sign is curious, certainly they shouldn't all give the same value of the correlation. They are entirely different shapes. HowiAuckland (talk) 01:35, 8 October 2008 (UTC)

OK, then take Tango's idea but use the cosine of the angle (that is, the dot product of the normalization of the sum of the vectors for each shape): ${\vec {X}}:=\sum \nolimits _{i}{\vec {x}}_{i}$, $c(\{{\vec {x}}\},\{{\vec {y}}\}):={\hat {X}}\cdot {\hat {Y}}$. Then rotating a shape by any angle will give a correlation with the original equal to the angle's cosine. It's still undefined if either of the shapes is regular, but you can perturb the angles slightly and get a function (of the 12 scalars) that's always defined but is very sensitive near the line $s_{1}=s_{2}=\dots =s_{12}$. You could also consider the (cosine of the) angle between the centroids of the two shapes; usually these are nearly the same thing, of course, so I don't know how to tell which one is better for your purposes. --Tardis (talk) 16:54, 8 October 2008 (UTC)

It's a good idea, but we are looking for a formula which works in all cases. As you have pointed out, in certain limits , i.e. regular, or small sums of vectors which are perturbed, the procedure is undefined, or not well defined or doesn't give answers that you should expect, such as +1. We had actually tried this idea previously and saw the defects of the procedure. Thanks so much for your opinions in this matter.HowiAuckland (talk) 22:21, 8 October 2008 (UTC)

I've been coming to the opinion, as CiaPan indicated above, that such a finalized formula for a single real number between −1 and +1, which works for all cases is most likely impossible to generate. I have been thinking that in order to study the structure of such a system, at least two numbers of correlation should be defined. A measure of asymmetry $(-1\leq c_{1}\leq +1)$ and a measure of similarity $(0\leq c_{2}\leq +1)$. I know how to determine a measure of similarity. I extracted this from similar concepts in hyper-spherical coordinates: Consider an angle defined by the formula $\tan \alpha _{i}={\frac {y_{i}}{x_{i}}}$ (with $\alpha _{i}$ set to be 0 in both cases of either $x_{i}=0$ or $y_{i}=0$, and $\tan \alpha _{i}={\frac {x_{i}}{y_{i}}}$ is equally good). Now consider $c_{2}={\frac {1}{6}}\sum _{i=1}^{6}\sin ^{2}(2\alpha _{i}).$ This works very well for a coefficient of similarity, but this measure does not capture the concept of asymmetry. Now I just need a formula, which is well-defined in all limits, to compute $c_{1}$ a coefficient of asymmetry. However, there is still the possibility that I might have missed something in my analysis and that a single formula for a single number might be well-definable. I still don't know for sure. HowiAuckland (talk) 22:21, 8 October 2008 (UTC)

side, edge of a circle

what is the difference between or the meaning of side of a circle and edge of a circle... How many sides does a circle have? How many edges does a circle have?

116.75.181.164 (talk) 10:43, 7 October 2008 (UTC)

The boundary of a circle is neither called side nor edge, but circumference. So the short answer to your question is: zero sides and zero edges. However it does make some sense to say that the circle has an infinite number of sides, as it is approximated sufficiently well by a polygon having sufficiently many sides. Bo Jacoby (talk) 11:21, 7 October 2008 (UTC).

But then again, infinity is not a number. —Preceding unsigned comment added by 81.187.252.174 (talk) 11:54, 9 October 2008 (UTC)

The 1-D Heat Equation

Help! I don't understand this Heat Equation stuff.

I have a problem with a semi-infinite cylinder of metal on the positive x-axis with sides and face at x=0 insulated. The heat will therefore diffuse according to the 1-D heat equation. The initial temperature distribution is $u(x,0)=u_{0}\exp(-x/L)$, the end is insulated which indicates that $u_{x}=(0,t)=0$.

I'm trying to solve this using Green's function, extended to the infinite axis as an even function, and then changing variables to get the error function. But I have been given the solution and no matter what I try I can't get to it. I don't think I understand exactly how to incorporate the initial value function as part of the integrand.

The solution given is $u(x,t)=(1/2)u_{0}exp(c^{2}t/L^{2})[exp(-x/L{1+erf((x-2c^{2}t/L)/{\sqrt {(}}4c^{2}t)}+exp(x/L){1-erf((x+2c^{2}t/L)/{\sqrt {4c^{2}t}}}]$

If anyone can explain where this solution comes from I would really appreciate it. —Preceding unsigned comment added by 122.108.235.248 (talk) 13:49, 7 October 2008 (UTC)

I suppose you mean $u_{x}(0,t)=0$. The c is unexplained. There are some missing parentheses in the formula which might have been

$u(x,t)=(1/2)u_{0}\exp(c^{2}t/L^{2})[\exp(-x/L)(1+\operatorname {erf} ((x-2c^{2}t/L)/{\sqrt {4c^{2}t}}))+\exp(x/L)(1-\operatorname {erf} ((x+2c^{2}t/L)/{\sqrt {4c^{2}t}}))]$

where

$\operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}dt$

is the error function.

Simplify the computations by choosing the unit of temperature to be $u_{0}$, and the unit of length to be $L$, and the unit of time such that $c=1$. Then the formula looks like this:

$u(x,t)=e^{t}[e^{-x}(1+\operatorname {erf} ((x-2t)/{\sqrt {4t}}))+e^{x}(1-\operatorname {erf} ((x+2t)/{\sqrt {4t}}))]/2$.

The initial condition at t→0⁺ is

${\begin{aligned}u(x,0^{+})&=[e^{-x}(1+\operatorname {erf} (x/0^{+}))+e^{x}(1-\operatorname {erf} (x/0^{+}))]/2\\&=[e^{-x}(1+\operatorname {erf} (+\infty ))+e^{x}(1-\operatorname {erf} (+\infty ))]/2\\&=[e^{-x}(1+1)+e^{x}(1-1)]/2\\&=e^{-x}.\end{aligned}}$.

OK! Then check the other initial condition

$u_{x}(0,t)=0\,$

and the heat equation.

$u_{t}-u_{xx}=0\,$

Bo Jacoby (talk) 20:02, 7 October 2008 (UTC).

Continuing. The u-function can be written compactly

$u={\frac {u_{+}+u_{-}}{2}}\,$

where $u_{+}$ and $u_{-}$ are defined by

$u_{\pm }=e^{t\mp x}\cdot (1\pm \operatorname {erf} (z_{\pm }))$

and $z_{+}$ and $z_{-}$are defined by

$z_{\pm }=t^{-{\frac {1}{2}}}({\frac {x}{2}}\mp t).$

The differential is

$du={\frac {du_{+}+du_{-}}{2}}$

where

${\begin{aligned}du_{\pm }&=d(e^{t\mp x}\cdot (1\pm \operatorname {erf} (z_{\pm })))\\&=d(e^{t\mp x})\cdot (1\pm \operatorname {erf} (z_{\pm }))+e^{t\mp x}\cdot d(1\pm \operatorname {erf} (z_{\pm }))\\&=e^{t\mp x}\cdot (1\pm \operatorname {erf} (z_{\pm }))\cdot d(t\mp x)\pm e^{t\mp x}\cdot 2\cdot \pi ^{-{\frac {1}{2}}}\cdot e^{-z_{\pm }^{2}}\cdot dz_{\pm }\\&=u_{\pm }\cdot (dt\mp dx)\pm e^{t\mp x}\cdot 2\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}-t\pm x}\cdot ({\frac {1}{2}}t^{-{\frac {1}{2}}}dx-{\frac {1}{4}}xt^{-{\frac {3}{2}}}dt\mp {\frac {1}{2}}t^{-{\frac {1}{2}}}dt)\\&=u_{\pm }dt\mp u_{\pm }dx\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}\cdot (t^{-{\frac {1}{2}}}dx-{\frac {1}{2}}xt^{-{\frac {3}{2}}}dt\mp t^{-{\frac {1}{2}}}dt)\\&=(\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})dx+(u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (-{\frac {x}{2}}\mp t))dt\\&=\mp (u_{\pm }-\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})dx+(u_{\pm }+\pi ^{-{\frac {1}{2}}}e^{-{\tfrac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (\mp {\frac {x}{2}}-t))dt\end{aligned}}$

This means that

${\frac {\partial u_{\pm }}{\partial x}}=\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}$

${\frac {\partial u_{\pm }}{\partial t}}=u_{\pm }+\pi ^{-{\frac {1}{2}}}e^{-{\tfrac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (\mp {\frac {x}{2}}-t)$

The initial condition

$u_{x}(0,t)=0\,$

is checked:

${\begin{aligned}u_{x}(x,t)&={\frac {1}{2}}\left({\frac {\partial u_{+}}{\partial x}}+{\frac {\partial u_{-}}{\partial x}}\right)\\&={\frac {1}{2}}\left(-u_{+}+\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}+u_{-}-\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}\right)\\&={\frac {1}{2}}\left(-u_{+}+u_{-}\right)\\&={\frac {1}{2}}\left(-e^{t-x}\cdot (1+\operatorname {erf} (z_{+}))+e^{t+x}\cdot (1-\operatorname {erf} (z_{-}))\right)\\&={\frac {e^{t}}{2}}\left(-e^{-x}\cdot (1+\operatorname {erf} (t^{-{\frac {1}{2}}}({\frac {x}{2}}-t)))+e^{+x}\cdot (1-\operatorname {erf} (t^{-{\frac {1}{2}}}({\frac {x}{2}}+t)))\right)\end{aligned}}$

The limit for $x\rightarrow 0$ is

${\begin{aligned}u_{x}(0,t)&={\frac {e^{t}}{2}}\left(-(1+\operatorname {erf} (t^{-{\frac {1}{2}}}(-t)))+(1-\operatorname {erf} (t^{-{\frac {1}{2}}}(+t)))\right)\\&={\frac {e^{t}}{2}}\left(-\operatorname {erf} (-t^{\frac {1}{2}})-\operatorname {erf} (t^{\frac {1}{2}})\right)\\&=0\end{aligned}}$

because

${\begin{aligned}\operatorname {erf} (-z)=-\operatorname {erf} (z)\end{aligned}}$

So far I have shown you that the u-function satisfies the two boundary conditions. It remains to be shown that it also satisfies the heat equation. Some other day, perhaps. Is this helpful at all? Bo Jacoby (talk) 20:08, 9 October 2008 (UTC).

Here we go again. A second order partial derivative has to be computed.

${\begin{aligned}{\frac {\partial ^{2}u_{\pm }}{\partial x^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial u_{\pm }}{\partial x}}\\&={\frac {\partial }{\partial x}}(\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})\\&=\mp {\frac {\partial {u_{\pm }}}{\partial x}}\pm \pi ^{-{\frac {1}{2}}}t^{-{\frac {1}{2}}}{\frac {\partial {e^{-{\frac {x^{2}}{4t}}}}}{\partial x}}\\&=\mp (\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})\pm \pi ^{-{\frac {1}{2}}}t^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}{\frac {\partial {-{\frac {x^{2}}{4t}}}}{\partial x}}\\&=u_{\pm }-\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}\mp {\frac {1}{2}}\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}x\\&=u_{\pm }+\pi ^{-{\frac {1}{2}}}e^{-{\tfrac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (\mp {\frac {x}{2}}-t)\\&={\frac {\partial u_{\pm }}{\partial t}}\end{aligned}}$

So $u_{+}$ and $u_{-}$ each satisfies the heat equation, which is linear, and therefore the linear combination

$u={\frac {u_{+}+u_{-}}{2}}\,$

also satisfies the heat equation

${\frac {\partial ^{2}u}{\partial x^{2}}}={\frac {\partial u}{\partial t}}$

I hope this answered your question. I got routine in writing wikipedia formulas. Bo Jacoby (talk) 15:40, 10 October 2008 (UTC).

Simple statistics question

(related to Talk:Mongolia): Assuming that 800 children are born in a hospital in one year, and assuming that the probability that a newborn is a boy is 0.5 (feel free to calculate with a more correct value, but then tell me so), what is the probability of more than 430 of the newborn being girls?

My impression is that this should be roughly 0.025 (normal distribution, E=400, sigma square = 200). Anything wrong about this? Yaan (talk) 18:50, 7 October 2008 (UTC)

Using a normal approximation should get a pretty good answer for n=800 (I haven't checked to see if you calculated it right, but the method is good). You could also do it directly as a binomial distribution (that article should have the formulae you need). The normal approximation is easier, though, if you don't need an absolutely precise answer. --Tango (talk) 18:59, 7 October 2008 (UTC)

From human sex ratio, "The natural sex ratio at birth is estimated to be close to 1.1 males/female", so you need to take that into account. But your stat is the reverse!? Saintrain (talk) 19:48, 7 October 2008 (UTC)

Yes, but the numbers are quite low, so it's actually not that unlikely of you repeat the "experiment" over a number of years, or a number of hospitals. The actual numbers from a statistical yearbook were 395 boys vs. 424 girls, i.e. a deviation from the mean of 14.5 if you use a sex ratio of 1. Standard deviation would then be 14.31. Though if you assume a ratio of 1.05 (p("boy")=0,512), the deviation would be 24.5, with the standard deviation slightly lower than before, and if you assume a ratio of 1.1 (p("boy")=0,524), the deviation from the mean would be 34, and the standard deviation only around 14.2. Yaan (talk) 14:43, 8 October 2008 (UTC)

The standard deviation is 200^1/2=14.1421. The deviation of the observation from the mean is 430−400=30. And 30/14.1421=2.12133. The probability that a normally distributed observation is greater than the mean + 2.12133 standard deviations is 0.016947, or roughly 1/59, (as compared to 0.025=1/40). Bo Jacoby (talk) 21:17, 7 October 2008 (UTC).

If you're going to claim so much accuracy, you should do a continuity correction using the fact that in this context, "> 430" is the same as "≥ 430.5". So

${\frac {430.5-400}{\sqrt {200}}}=2.1567,$

and the probability that the standard normal exceeds that is about 0.015515, or about 1/64. Michael Hardy (talk) 01:08, 8 October 2008 (UTC)

You are absolutely right! Do you know the binomial probability without the error due to the normal distribution approximation? Bo Jacoby (talk) 05:51, 8 October 2008 (UTC).

The exact answer is given by the binomial distribution. The R idiom would be sum(dbinom(431:800 , 800 ,prob=0.5)), giving 0.01548346. Robinh 11:13, 8 October 2008 (UTC) [what is the correct markup for verbatim?]

Try pre or code tags, one of them should do what you want. You may also require nowiki tags. I've never quite got my head round the different combinations of those tags. --Tango (talk) 12:50, 8 October 2008 (UTC)

Thanks a lot everybody. Yaan (talk) 14:43, 8 October 2008 (UTC)