[Lign251] HW3 Q3

[Kathryn Davidson]

In problem 3, I have a perfectly good plot (using plot()) of the function
that seems to estimate the histogram. However, I am really having problems
using lines() to overlay it on top of the histogram. It just keeps saying:

Error in as.vector(x, "double") : cannot coerce to vector

So I guess my function isn't in vector format... but I can't figure out how
to make it one. In Baayen pg. 27 I think he's answering this question but I
can't follow what he's saying.
Any tips?
Kate

On 10/21/07, rlevy at ling.ucsd.edu <rlevy at ling.ucsd.edu> wrote:
>
> Rebecca Colavin wrote:
> > Okay, I give up.
> >
> > The graph in the question generates a distribution where the probability
>
> > mass (=1) is contained under the curve. In short, the y axis is (0,9) or
>
> > so .
> >
> > But when I try to do lines over it with a normal distribution, the y
> values are probabilities. The y axis for this is (0,1).
>
> Hi Rebecca,
>
> Because the normal distribution is a continuous probability density, the
> y values (for the density) are not bounded above by 1.  Remember, it is
> the *area under the curve* that must total 1.  If the scale of the x  axis
> is small, p(x) can easily surpass 1.
>
> As an example, try:
>
> > x <- seq(-5,5,by=0.01)
> > plot(x,dnorm(x,0,0.1),type="l")
>
> This normal density peaks around p(x) = 4.
>
> Using hist() with prob=T also is an estimation of density, which is why
> the hist() y axis also surpasses 1.
>
>
> Roger
>
>
> --
>
> Roger Levy                      Email: rlevy at ucsd.edu
> Assistant Professor             Phone: 858-534-7219
> Department of Linguistics       Fax:   858-534-4789
> UC San Diego                    Web:   http://ling.ucsd.edu/~rlevy
>
>
>
>
>
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