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10. I have a data set which displays a bimodal distribution. This was determined by plotting a histogram of the frequency vs number. I now need to separate the two original populations and therefore find an intersection point of sorts. From the plot it looks like the point might be approx. -1.0 to -0.8.
I have some bimodal data like the one generated down (R language), and I don't know how to transform it to have a normal distribution or homoscedasticity. I'm running a linear discriminant analysis and I need homoscedasticity, but I'm not able to get it with this kind of distribution.
3. If the histogram were actually the distribution that the data were drawn from (it would then be a piecewise uniform one, clearly), you could say it was right skew (by pretty much any reasonable measure) and multimodal, since there's clearly more than two modes. But presumably we're trying to use the histogram to infer something about the ...
If the data comes from daily observations of temperature T T on day d d of the year over a year or two, you could report the coefficients for a best fit of the model. T = a + b cos(d 3652π) T = a + b cos. . (d 365 2 π) or perhaps more precisely. T = a + b cos(d + 10 365 2π) T = a + b cos. .
A simple way to program a bimodal distrubiton is with two seperate normal distributions centered differently. This creates two peaks or what wiki calls modes. You can actually use almost any two distributions, but one of the harder statistical opportunities is to find how the data set was formed after combining the two random data distributions.
We can re-frame this problem in a different way. It is a very common problem to find a threshold to distinguish two classes in a bimodal distribution like this. Even in image processing and computer vision it is a common problem to find a threshold in a bimodal distribution to make a grayscale image to a binary image.
The Central Limit Theorem works for bimodal distributions. To verify that averages of samples as large as ours tend to be normal, we can re-sample from x1. Ten thousand averages, re-sampled (with replacement) of size 3000, are nearly normally distributed as shown in the histogram below. The red normal curve is a reasonably good fit to the ...
If we run a model without including sex, so just height ~ weight then we get the following results. "height"=c(rnorm(100,160,3),rnorm(100,180,3)), "weight"=rnorm(200,70,10), "sex"=factor(c(rep("F",100),rep("M",100))) The reason being that a linear model will fit the mean regression line, which happens to be right in the middle of F and M, which ...
Beware, however -- histograms can have problems, too; indeed, we see one of its problems here, because the distribution in the third "peaked" histogram is actually distinctly bimodal; the histogram bin width is simply too wide to show it. Further, as Nick Cox points out in comments, kernel density estimates may also affect the impression of the ...
1. Standard deviation, whilst it is often used on normal distributions, is it a useful statistic on global temperature, both daily and yearly, given that weight of the data is towards the ends of the histograms i.e. Bimodal? This means that a very small multiple of any SD (say 1.5) will cover 100% of the data. Example data is the following: