BruceW said:'variance' is kinda strange given just the problem statement... I guess it means the moment of inertia about the centre of mass, when the total mass equals 1 ?
But then in your attempt, you use a probability distribution ##3/4\pi R^3## Where did this come from? This wasn't in the problem statement.
BruceW said:what is your pdf supposed to be? was it specified?
BruceW said:hmm, interesting. what do you think they wanted? for the mean, you could calculate the centre of mass, where the density is only nonzero (and is constant) in the specified region. For the variance, it's not so simple. maybe use squared distance from the centre of mass? Also, for this problem, are you using all the points within the ball of some radius? Or all the points on the surface of this ball?
which implies to me that they want you to consider the points only on the surface of the sphere. But now you are saying they want you to consider all the points within the sphere? well, I guess you just need to choose which one you think is most likely. Also, for 'variance of a point' I know what equation I would guess this means. But I don't really know. You should try to figure out what equation they want you to use. What would the variance be in 1d ? and try to extend that to something that is a similar equation in 3d.unscientific said:Consider the sphere x2 + y2 + z2 = 1
Find the mean and variance.
BruceW said:In the first post, you wrote
which implies to me that they want you to consider the points only on the surface of the sphere. But now you are saying they want you to consider all the points within the sphere? well, I guess you just need to choose which one you think is most likely. Also, for 'variance of a point' I know what equation I would guess this means. But I don't really know. You should try to figure out what equation they want you to use. What would the variance be in 1d ? and try to extend that to something that is a similar equation in 3d.
BruceW said:ah, that's much clearer! OK, since they say 'randomly chosen within the sphere', this implicitly tells us what the PDF is. As another example, if I said I will randomly choose a number between 0 and a given number ##L##, then what is the PDF in this case? That is the 1d case, so then it is straightforward to go to the 3d case.
BruceW said:almost right. The pdf should be a constant within the specified area or length or volume or whatever. But also, when you integrate the pdf over that total area, or length or volume, you should get 1 (since you will certainly choose a point, out of the possible options). So this means the pdf of numbers distributed from 0 to 10 is definitely not 5. and from 0 to 1 is not 1/2.
BruceW said:hmm. your teacher should have explained stuff like this before giving this kind of problem. Yes, the pdf is constant within the specified volume, and zero outside. And by the definition of a pdf, you know the volume integral of the pdf must equal 1. So le'ts call the constant ##C##. Then:
##\int_{volume} C \ dV = 1##
where the volume integral is only over the specified volume, since the pdf is zero outside of the volume. And now, you know ##C## is a constant, so what is the next step?
BruceW said:yep, that's it! And be sure to keep in mind that ##r## in those equations is a constant, not a variable. i.e. for the volume, ##r## is the total radius of the sphere (a constant). Looking at your first post, I think you got it partly correct. They are asking for the mean distance from the origin. Not the mean position from the origin. So the mean will not be zero. Now you have the pdf, you can find the mean of the distance from the origin in the usual way.
The average of a unit sphere is the center point, or centroid, of the sphere. It is the point at which all points on the surface of the sphere are equidistant from.
The variance of a unit sphere is calculated by finding the average distance of all points on the surface of the sphere from the centroid. This distance is then squared and divided by the surface area of the sphere.
The average and variance of a unit sphere are both measures of central tendency and variability, respectively. They are related in that the variance is a measure of how spread out the points are from the average, or centroid, of the sphere.
The dimension of a unit sphere does not affect its average, as the centroid will always be located at the center point. However, the dimension can affect the variance, as a higher dimensional sphere will have a larger surface area and therefore a larger variance.
Yes, the average and variance can be used to describe the distribution of points on the surface of a unit sphere. The average gives the center point, while the variance gives an indication of how spread out the points are from that center point.