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November 18[edit]

statistics[edit]

Hi, I'm just reviewing an old statistics problem, and it says the solution to one of the problems is this:

A machine can be set to discharge soil in amounts that are normally distributed with mean μ tonnes and standard deviation equal to 25.7 tonnes. If the farmer wishes to use the machine to fill containers that hold 2000 tonnes of soil and wants to overfill only one container in 100, at what value of μ should the farmer set the machine?

Answer: Let X be the amount of soil. X is normal with mean μ, σ = 25.7 and Z = (X-μ)= σ is a standard normal random variable. We wish to determine μ such that P(X > 2000) = .01. This will occur if P (Z > 2000 -μ /25.7) = .01

Then it says: Using the z-table, we find that z_.01(is nearly equal to) 2.33

But the z table doesn't have 2.33 on it! Could someone please clarify this for me? I can type out the rest of the question if you need more information. --Jeevies (talk) 10:14, 18 November 2008 (UTC)[reply]

My normal distribution table gives that value. What table are you using? Algebraist 11:34, 18 November 2008 (UTC)[reply]

you do know how to read the table right? You get the first 2dp from the column on the left and the third dp from the row along the top. So even though there is no value exlpicitly on the column for 2.33 this information is in the table if you know how to read it. —Preceding unsigned comment added by 129.67.39.18 (talk) 13:15, 18 November 2008 (UTC)[reply]

He/she may need to use interpolation to find the value for 2.33 and not the value for 2.3. I don't remember the exact process at the moment so if this question isn't answered in a while I'll try to find my statistics book to help out.--droptone (talk) 15:37, 18 November 2008 (UTC)[reply]

Just using a better table is a lot easier. Algebraist 18:53, 18 November 2008 (UTC)[reply]

I'm sorry, I meant to say that I don't understand how to obtain 2.33 from 0.1 --Jeevies (talk) 19:09, 18 November 2008 (UTC)[reply]

The table Algebraist referred to gives approx. 0.99 left area, 0.01 right area, for z=2.33; you find the value, in whichever form of the table suits your problem, closest to the one required then read off the z value.86.155.186.16 (talk) 19:39, 18 November 2008 (UTC)[reply]

No one uses tables nowadays. Just that species of animal: a graphic calculator.

Topology Expert (talk) 04:03, 19 November 2008 (UTC)[reply]

Could you clarify a bit more please? It's still a mystery why 2.33 was picked out of all the (approx) 0.01 values. --Jeevies (talk) 16:33, 19 November 2008 (UTC)[reply]

To use Algebraist's table: The table is of the probability that something is less than Z. You want 0.01 probability that it is larger than Z, so 0.99 probability that it is less than Z. Find the value in the table that is closest to 0.99. It is 0.9901, located on the row labeled 2.3 in the column labeled 0.03. So your desired Z is 2.3 + 0.03. -- Jao (talk) 21:03, 19 November 2008 (UTC)[reply]

Vector Perdipendicular Product[edit]

One day while I was thinking about vector cross products I realized you could have a function that intakes (n-(sgn(n))) vectors and output a n dimensional vector perdipendicular to all the inputed vectors. It would behave like cross products with two vectors. I tried solving it with a Matrix equation, and the vector would be the transpose of x is the vector product if the matrix equation is Ax=0. But in solving it, you need to find the transpose of (0A) which is the zero matrix, which is wrong! So can anyone tell me where did I go wrong and how do I correct it? ----~~The Successor of Physics~~ 14:09, 18 November 2008 (UTC)—Preceding unsigned comment added by Superwj5 (talk • contribs) 14:07, 18 November 2008

I don't really know, but while you're waiting for someone else to come along, try reading Exterior algebra. I think you may find something useful there (the wedge product seems to do something close to what you're asking for). --Tango (talk) 14:32, 18 November 2008 (UTC)[reply]

This corresponds to finding the null space of the

(n-1)\times n

matrix whose rows are your n-vectors. It's a linear operator (with an arbitrary scale) on any one vector if you fix the others. Also, spelling: perpendicular. --Tardis (talk) 17:20, 18 November 2008 (UTC)[reply]

gauss?[edit]

I remember something I came across a few years ago.. it was something like you can calculate the sum of every other number, up to some number, with a square of something.. does anyone know what I'm talking about? .froth. (talk) 23:26, 18 November 2008 (UTC)[reply]

If n is odd, then the sum of all odd numbers from 1 to n is ((n+1)/2)². Is that what you're talking about? Algebraist 23:33, 18 November 2008 (UTC)[reply]

Square number#Properties makes this easy to see with graphics. PrimeHunter (talk) 00:01, 19 November 2008 (UTC)[reply]

Yeah I think it was that one. All squares are the sum of a series of odd numbers from 1 to some n -froth —Preceding unsigned comment added by 71.176.164.189 (talk) 22:23, 19 November 2008 (UTC)[reply]

Perhaps you're thinking of the formula for the triangular numbers? It isn't precisely a square, but it does involve the square of the upper limit of the sum. Gauss is supposed to have noticed this in elementary school. --Tardis (talk) 23:58, 18 November 2008 (UTC)[reply]