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April 5

Vector proof

Given a Quadrilateral $ABCD$ and $E,F$ are midpoints of $AD,BC$ respectively,

Prove: ${\overrightarrow {EF}}={\frac {{\overrightarrow {AB}}+{\overrightarrow {DC}}}{2}}$ . יהודה שמחה ולדמן (talk) 09:44, 5 April 2016 (UTC)

Please see our note at the top of the page regarding homework. What have you done so far and where are you stuck?

To start you out:

• Have you drawn a diagram of the problem?
• Do you know and understand why, given the points $W,X,Y,Z$, then ${\overrightarrow {WZ}}={\overrightarrow {WX}}+{\overrightarrow {XY}}+{\overrightarrow {YZ}}$?
• How many ways can you represent ${\overrightarrow {EF}}$ in terms of vectors not utilizing the points $E,F$?
• Can you think of a way to combine those equations so that you can solve for ${\overrightarrow {EF}}$ in the desired terms?

Our relevant article is Vector addition. -- ToE 14:07, 5 April 2016 (UTC)

April 6

Sums using multiple variables: A question about sigma notation

Hello, I have a question about sigma notation. One could write the following, having the understood meaning shown afterward in an equality: $\sum _{i=1}^{4}{\binom {7}{i}}={\binom {7}{1}}+{\binom {7}{2}}+{\binom {7}{3}}+{\binom {7}{4}}$

However, what if we have two variables, i and j, with i going UP from 1 to 3 and j going down from 4 to 2, simultaneously? (NOT a double sigma where all possibilities of i,j are summed!) I want to express the following sum: ${\binom {1}{4}}+{\binom {2}{3}}+{\binom {3}{2}}$

I am tempted to write as follows, or similar: $\sum _{i=1,j=4}^{i=3,j=2}{\binom {i}{j}}$ OR $\sum _{i=1,j=4}^{3,2}{\binom {i}{j}}$ etc...

... but this seems silly! How can i PROPERLY express what i am trying to, using sigma notation? (Or a better way, if what im trying to do as short hand isnt appropriate)

216.173.144.188 (talk) 04:41, 6 April 2016 (UTC)

One option is $\sum _{i=1}^{3}{\binom {i}{5-i}}$. -- ToE 06:04, 6 April 2016 (UTC)

Yes, I think that's the way it would usually be done. If both i and j were used a lot of times in the expression, I suppose you could write something like

$\sum _{i=1}^{3}{\binom {i}{j}}$ where $j=5-i$

and it would be understood. --69.159.61.172 (talk) 06:47, 6 April 2016 (UTC)

Okay, thank you, this makes sense! However, it is a simplification based on what i guess now i see as a poor substitute example. Let me ask about my actual situation and see what the correct way is to write this concisely as a mathematician. We have the following:

Given $n\in \mathbb {N} ^{+}$, Consider $\sum _{j=\lfloor {\frac {n}{2}}\rfloor }^{0}{\binom {n-j}{j}}$ ...

Does this make sense? Is there any reason why the index of a sum cant decrease instead of increase?

(It took me a while to realize i could express i in terms of n and j together, which helped in this)

216.173.144.188 (talk) 13:20, 6 April 2016 (UTC)

In the sigma notation the index can't decrease because Summation#Formal_definition says that $\sum _{i=a}^{b}g(i)=0$ $\,$, for b < a. The fix is simple though, just write $\sum _{j=0}^{\lfloor {\frac {n}{2}}\rfloor }{\binom {n-j}{j}}$. Egnau (talk) 16:02, 6 April 2016 (UTC)

Indeed,

Thank you all very much for your input. I consider this matter closed. 216.173.144.188 (talk) 16:35, 6 April 2016 (UTC)

The Iverson bracket is useful.

$\sum _{i=1}^{3}\sum _{j=0}^{\infty }[i+j=5]{\binom {i}{j}}$

Bo Jacoby (talk) 22:31, 6 April 2016 (UTC).

What are the odds that a person will die on his birthday?

What are the odds that a person will die on his birthday? Is it 1 out of 365? Or is it more complicated than that? I was surprised to see that this happened with Merle Haggard today. Thanks. Joseph A. Spadaro (talk) 19:49, 6 April 2016 (UTC)

yes, would have to be, if looking at whole population...or whatever it end up being if include the extra leap days....UNLESS it can be determined that more people tend to be born in the spring or summer and more people tend to die in the fall or winter.....68.48.241.158 (talk) 20:26, 6 April 2016 (UTC)

But once you have a birthday, you have a birthday. That's that. So, it's simply a matter of "matching" the death day. No? Joseph A. Spadaro (talk) 20:28, 6 April 2016 (UTC)

Close to 365, unless they were born on leap day. I suppose a surprise and big birthday meal might put the death rate ever so slightly higher that day. StuRat (talk) 20:27, 6 April 2016 (UTC)

if 65% of population is born in spring or summer and 65% of population dies in fall or winter, then looking at the whole population...a human being has a smaller chance than one 1/365(or so) of dying on her birthday... 68.48.241.158 (talk) 20:34, 6 April 2016 (UTC)

I am asking about a specific person, not just any generic person. And a specific person, once born, already has a set birth date (whatever that date might be). So, the question is not so much: "What are the odds that any one of us human beings will die on our birthday?" The question is more like: "What are the odds that Brad Pitt will die on his birthday?" Which is another way of saying: "What are the odds that Brad Pitt will die on December 18?". Joseph A. Spadaro (talk) 21:14, 6 April 2016 (UTC)

if asking specific person than depends on individual factors..so unanswerable...person might be dying of cancer, only have 6 weeks to live...birthday 6 months away.....?????68.48.241.158 (talk) 21:21, 6 April 2016 (UTC)

but if brad pitt is expected to live many, many more years (and human being are more likely to die in fall or winter) then he has a smaller chance of dying on his birthday....68.48.241.158 (talk) 21:23, 6 April 2016 (UTC)

... if he is living in the southern hemisphere? Dbfirs 21:29, 6 April 2016 (UTC)

^right, my mistake...general reasoning okay though, I think...68.48.241.158 (talk) 21:35, 6 April 2016 (UTC)

Actually, is there a mistake? December 18 is still technically fall, I believe (not winter). I think winter starts on December 21 or so. No? Joseph A. Spadaro (talk) 22:29, 6 April 2016 (UTC)

There are many definitions of when seasons start. Don't be fooled by news people (especially in North America) who like to talk about "the official first day of winter". Besides, everyone knows the official first day of winter is in October. --69.159.61.172 (talk) 23:07, 6 April 2016 (UTC)

Summer starts on December 1st in almost all of the temperate southern hemisphere, though I expect someone will be able to find an odd exception. Dbfirs 07:46, 7 April 2016 (UTC)

Agreed. There are slightly more deaths in winter where I live and also where you live, but that will not be the case in some other parts of the world. Dbfirs 21:50, 6 April 2016 (UTC)

That would be true if seasonal variation were as great in humans as it is in some other animals. The answer will be approximately "364:1 against" (the question asked for odds, not probability), but with slight variations as mentioned above. Dbfirs 21:06, 6 April 2016 (UTC)

certainly there's some variation so determining the actual odds or probability (is there a difference in this context?) would in practice be complicated...68.48.241.158 (talk) 21:14, 6 April 2016 (UTC)

Yes, I wasn't disagreeing, just commenting that the effect would be small. It would also vary by region and culture. I agree that the fine detail would be complicated to calculate. In the future, it might be possible to predict, from an individual's DNA, address, and lifestyle, the most likely time of year of death, but the probability is unlikely to change significantly from 1/365 (the small correction for leap years possibly cancels out the small birthday surprise effect or birthday parachute jump). The answer for Brad Pitt will be very close to the answer for any other individual. Dbfirs 21:26, 6 April 2016 (UTC)

(Not to be morbid, but it's a morbid question anyway) Infant mortality may dominate other effects. 11000 / 4000000 ≈ 1/350 of all babies born in the U.S. die within 24 hours of birth. Probably more than 1/700 die on the same date (more births are in the morning, and probably more deaths are earlier in the 24-hour period too). That's quite large compared to 1/365. -- BenRG (talk) 04:55, 7 April 2016 (UTC)

Good point. We should probably restate the problem as "What is the probability of dying on your birthday, but not your birth date". StuRat (talk) 05:17, 7 April 2016 (UTC)

In that case, mortality is highest on the day after the birthday, and still high the days after that, from infant deaths with a small delay after birth. Then, deaths on the birthday itself are probably even lower than average since it is the last date to appear after we start counting, and therefore least affected by infant mortality. Unless some other intrinsic birthday effect offsets this, of course. If all effects related to infant mortality are to be excluded from the question, one should exclude all data below age 5 or so. Gap9551 (talk) 16:18, 8 April 2016 (UTC)

Yes, I apologise to all new-born babies for not considering them as "person"s. Dbfirs 07:56, 7 April 2016 (UTC)

• Check out "Variation of mortality rate during the individual annual cycle", which investigates all births/deaths in the city of Kiev over the period 1990-2000. Kiev men showed a 44.4% excess in deaths on their birthday compared to the expected value (women slightly less at 36.2% excess), and it looks like other studies have found similar results. So working out the exact odds would take a lot of actuary work, but it looks like it's something more like 1:300 than 1:365. Smurrayinchester 08:46, 7 April 2016 (UTC)

I've started a page Birthday effect which collects the studies. Please feel free to improve it. Smurrayinchester 11:53, 7 April 2016 (UTC)

Thank you for that excellent summary of the research. The effect is much bigger than I expected. Dbfirs 14:14, 7 April 2016 (UTC)

Very interesting. I will have to read that new article. Thanks. Joseph A. Spadaro (talk) 18:04, 7 April 2016 (UTC)

I downloaded the first third of the Social Security Death Index (2.9 GB). From the 28,607,398 recorded birth / death pairs, I found 45,707 individuals who were recorded as dying on their birthday (0.1597%) giving a chance of approximately 1 in 625 of dying on your birthday (well below the expected 1 in 365). It should be noted that the SSDI only records deaths of individuals who were issued US social security numbers. As SSNs are not issued to individuals who die prior to the birth being officially registered, none of the records include infants who died on the day they were born. Dragons flight (talk) 18:50, 7 April 2016 (UTC)

As for seasonal differences:

	January	February	March	April	May	June	July	August	September	October	November	December
Births	2420302	2269663	2525845	2330376	2386245	2296193	2442604	2487487	2457619	2387806	2226876	2376382
Deaths	2708214	2407497	2539914	2348925	2331869	2217327	2290327	2248835	2197259	2391209	2362526	2563496

Dragons flight (talk) 18:59, 7 April 2016 (UTC)

Same dataset has 2,393,596 individuals (8.37%) dying in the same month that they were born, or a chance of 1 in 11.952. Dragons flight (talk) 19:10, 7 April 2016 (UTC)

Why would there be such a disparity between what we would expect (statistically) and the actual results? That does not seem to make sense. I am referring to this comment: I found 45,707 individuals who were recorded as dying on their birthday (0.1597%) giving a chance of approximately 1 in 625 of dying on your birthday (well below the expected 1 in 365). Thanks. Joseph A. Spadaro (talk) 21:00, 7 April 2016 (UTC)

What is the distribution of death dates? E.g., how many days (out of 365) have frequency less than 0.16% ? --JBL (talk) 21:09, 7 April 2016 (UTC)

Hhmmm, I've now taken a closer look at the data and it appears that about half of the SSDI death dates are recorded without a real day of death, e.g. March 0, 1979. Some of the birth dates also are also missing detail (but a much smaller fraction than the death dates). If I exclude all of the birth / death pairs that don't involve fully-formed dates, I am left with 14,879,058 individuals. Of these 44,665 were recorded as dying on their birthday (0.3002%; a chance of 1 in 330). That is now slightly more than the expected 1 in 365, though much closer than the initial estimate. Dragons flight (talk) 07:19, 8 April 2016 (UTC)

Looking a bit further:

Birthday + X days	Chance of dying (according to SSDI records)
-40 days	1 in 364
-30 days	1 in 366
-20 days	1 in 366
-10 days	1 in 366
-5 days	1 in 366
-4 days	1 in 366
-3 days	1 in 363
-2 days	1 in 370
-1 days	1 in 365
Birthday	1 in 330
+1 days	1 in 361
+2 days	1 in 361
+3 days	1 in 363
+4 days	1 in 364
+5 days	1 in 365
+10 days	1 in 363
+20 days	1 in 363
+30 days	1 in 365
+40 days	1 in 366

Dragons flight (talk) 08:08, 8 April 2016 (UTC)

that seems more in line with common sense...the previous estimate seemed weird..68.48.241.158 (talk) 11:54, 8 April 2016 (UTC)

confused about the chart though...saying only 1/364 chance dying the time frame of 40 days before birthday?? that can't be right...??68.48.241.158 (talk) 16:33, 8 April 2016 (UTC)

It's the chance of dying exactly that many days before or after your birthday. Dragons flight (talk) 17:00, 8 April 2016 (UTC)

That chart seems strange. All of the time intervals are listed with odds of very close to 1 in 365. They are all 362, 363, 364, 366, etc. Why would the exact birthday be the only one with a "weird" amount listed for odds as 330? That (the birthday) is the only one on that chart that really deviates from the expected 365 value. Why? Joseph A. Spadaro (talk) 18:25, 8 April 2016 (UTC)

Or does that strange value (330) include babies that die on the exact day they are born? But even that theory wouldn't make sense, I don't think? Joseph A. Spadaro (talk) 18:31, 8 April 2016 (UTC)

good question, as he said infants who died day they were born are not included...this would seem the only possibility that skew the numbers that drastically....68.48.241.158 (talk) 18:41, 8 April 2016 (UTC)

Yes. But doesn't the concept of "infant mortality" extend to the 1, 2, 3, 4, 5 or so days after the birth, also? Not just the exact birthday. So, those other surrounding days should also end up being skewed. No? Joseph A. Spadaro (talk) 19:13, 8 April 2016 (UTC)

Perhaps you should go reread the post in this thread in which Dragons flight explains where the data comes from. --JBL (talk) 20:15, 8 April 2016 (UTC)

well, how you explain the huge anomaly on the actual birthday???68.48.241.158 (talk) 20:20, 8 April 2016 (UTC)

Exactly! Joseph A. Spadaro (talk) 20:40, 8 April 2016 (UTC)

For a specific person - and estimating the odds on a specific day, there are more things to consider. Imagine, for example, two people who were born just two days apart - with identical lifestyle, identical current health, etc. Suppose A's birthday was yesterday, and B's birthday is tomorrow. If (for example) they both have some terminal disease with less than a month to live - then B can still die on his birthday - but A cannot. If they'll both live for less than two years, then B has two chances to die on his birthday - but A has only one. Of course if we do the estimation two days from now, their odds of having birthday deaths are very similar again. SteveBaker (talk) 18:55, 8 April 2016 (UTC)

not really relevant...yes, if looking at one specific person all kinds of things could come in play...but this is looking at odds for a large population as a whole...(you're absolutely right about what you say though...this was mentioned earlier in thread too)..68.48.241.158 (talk) 19:07, 8 April 2016 (UTC)

A couple of other things to consider, if people can find stats for them: 1) Placebo effect - could potentially result in people close to death and close to their next birthday managing to hang on until they reach it. 2) People celebrating their birthday by getting really, really drunk, and dying of alcohol-related accident or illness. Iapetus (talk) 10:09, 11 April 2016 (UTC)

April 7

Area, vertex, and boundary centroids

Two related questions: For what polygons does the area centroid (centroid of the entire enclosed region) coincide with (a) the centroid of the vertices, or (b) the centroid of the boundary points? For (a), it's true for all triangles and of course for all regular polygons. I'm guessing it's true for all parallelograms too. But what other polygons? As for (b), it does not hold for all triangles or even for all isosceles triangles; it holds for all regular polygons, and I'm guessing for all parallelograms, but what else? Loraof (talk) 19:04, 7 April 2016 (UTC)

Trivial remarks: obviously for any region with a line of symmetry, all three centroids must lie on that line. If there are two lines of symmetry, all three lie at the intersection and so coincide. (In this case there is a rotational symmetry, and so it is reasonable to speak of "the center" of the region.) --JBL (talk) 20:50, 7 April 2016 (UTC)

Case (a) is much easier since the area centroid and vertex centroid are both rational functions of the vertex coordinates. For the edge centroid you need to weight by the edge lengths, which gives you √'s in the expression and everything gets messier. For quadrilaterals I'm pretty sure (meaning I have most of a proof worked out but it's too much trouble to write out the deatials) that (a) is true iff it's a parallelogram. The crucial point is, and correct me if I'm wrong about this, that as operators on quadrilaterals, the point centroid and area centroid both commute with affine transformations. So you can simplify by taking three points of your quadrilateral to be (0, 1), (0, 0), and (1, 0). This idea won't work for the edge centroid because a linear transformation does not change length by a constant factor. --RDBury (talk) 03:30, 8 April 2016 (UTC)

Interpolation and extrapolation

I have been told, and told to teach, that interpolation is generally more reliable than extrapolation. Obviously, everything in science and statistics rests, at least tacitly, on a bed of assumptions; for example, in the case of science, we usually assume that induction works. What are the "minimal" statistical assumptions required for extrapolation to be generally less reliable than interpolation?--Leon (talk) 20:07, 7 April 2016 (UTC)

I don't think "interpolation is more reliable than extrapolation" is a formalizable statement in the sense you're hoping for. --JBL (talk) 20:52, 7 April 2016 (UTC)

From the lede of our article Extrapolation: "It is similar to interpolation, which produces estimates between known observations, but extrapolation is subject to greater uncertainty and a higher risk of producing meaningless results." -- ToE 21:14, 7 April 2016 (UTC)

@Thinking of England: If this is meant as a response to me then I don't understand its relevance to my comment. If it is meant only as a response to Leon, please feel free to remove or ignore this comment. --JBL (talk) 21:40, 7 April 2016 (UTC)

@Thinking of England: But that just tells me what I've already been told, with no explanation of why.

@Joel B. Lewis: In that case, is it in the same category as induction? By this I mean that it is a ill-defined heuristic that is important to science.--Leon (talk) 09:44, 8 April 2016 (UTC)

In my opinion, yes, this is a (well-founded) heuristic that one should expect to hold in any reasonable model, but not actually a statement with a fixed formal meaning. --JBL (talk) 20:21, 8 April 2016 (UTC)

(ec) I think it's something like this: if you have a simple regression line, I think you can calculate the variance of the predicted value (i.e., its variance over a large number of replications of the regression) as a function of the independent variable's value x. I think this variance is greater if you are outside the range of values of x that you used in the regression. The reason is this: draw a linear regression line through some data points. Suppose it correctly goes through a point in the middle, which I'll call the pivot point. Now except by coincidence, the estimated slope coefficient will not be exactly right; the correct line goes through the pivot point with a somewhat different slope. The farther away from the pivot point you look, the more the true line and your estimated line diverge from each other.

Another less formal way in which the maxim is true is this: a linear (or any other) regression may have its functional form misspecified. This might not matter much in the range that the x data used for the regression were in, but matters more the farther outside that range that you try to extrapolate. For instance, draw a bunch of data points almost exactly satisfying y=x². Pick a narrow range and run a linear regression. It may fit reasonably well for the data points included in the regression, but fits worse the farther outside that range you look. Loraof (talk) 21:20, 7 April 2016 (UTC)

Assuming only that uncertainty increases monotonically with distance outside a range of samples, there is no limit to the error that an excessive extrapolation can incur. See Extrapolation#Quality of extrapolation. AllBestFaith (talk) 02:04, 8 April 2016 (UTC)

I'd say it has to do with the assumption that the underlying function you're interpolating is continuous and "smooth" on the same scale as your point sampling density. That is, when interpolating, you're typically assuming the tightest radius of curvature for your function in the region you're sampling is on the same order or greater than the distance between any two sampled points. If those assumptions are true, you can't introduce substantial oscillations between any two points: the function is confined to a certain delta above/below the two points. Even with those minimal assumptions, though, extrapolation doesn't share that property. Even with a radius of curvature several times that of the point separation, you don't have to go very far before you can get a continuous function to go near-vertical. - However, violating either of those assumptions can allow interpolation to be as error-prone as extrapolation. Discontinuous functions can jump all over the place, and if your tightest radius of curvature is allowed to be even a quarter of the point separation distance, then you can get arbitrarily far from the points and still be able to return to fit both sides. -- 162.238.240.55 (talk) 15:01, 10 April 2016 (UTC)

I should add that even without radius of curvature issues (e.g. when you know the underlying functional form is linear, but with uncertain coefficients), extrapolation is still less accurate than interpolation, due to compounded errors of the slope. Within the range of your interpolated points, you know that any estimate should have an error ε, the error in fitting your points. However, your fitting protocol will have some error δ in its estimate of the slope. Even if δ is much less than ε, you get a δ of error for each unit you travel from the center. At some distance away from the points the sum of the δ's is going to be much greater than ε - if you go out far enough, arbitrarily so. -- 162.238.240.55 (talk) 15:18, 10 April 2016 (UTC)

I look at as how far you extrapolate or interpolate from the last data point. If we put it in terms of the distance between the last two data points, then when you interpolate, the farthest you can go is 50% of that distance. An extrapolation of 50% of that distance might not be too bad, either. On the other hand, you can extrapolate much further, but should then expect much worse results. StuRat (talk) 15:26, 10 April 2016 (UTC)

April 8

Goat problem

Several centuries ... I have been searching (and got stuck) in the 19th century:

Here (beautiful poem!)

... and here (3 horses).

(a) Is someone able to find older descriptions of this problem?

(b) Is the term "goat problem" the official/ultimate/best description - or are there better ones? THX 213.169.163.106 (talk) 11:45, 8 April 2016 (UTC)

I don't think either of those is the goat problem as described in Wikipedia. The first one just asks for the radius of a circle whose area is 1 acre (the answer is ${\sqrt {1{\mbox{ acre}}/\pi }}$). The second is a different puzzle involving regions bounded by circular arcs. -- BenRG (talk) 22:40, 8 April 2016 (UTC)

Then apart from the faulty examples: (a) (... centuries...) and (b) ? THX 213.169.163.106 (talk) 05:45, 9 April 2016 (UTC)

The AMM article listed as a reference also seems to have little to do with the problem stated in the WP article. It does give a problem dating from the 18th century, but that involves finding the area bounded by a circle and its involute. So there seems to some confusion all around as to which problem is "the goat problem". Mathworld lists both the our version and the involute in its article, but it's hard to tell from from it what is historical, what's modern and what's Weisstein's original research. --RDBury (talk) 13:08, 9 April 2016 (UTC)

This was my impression too. Any other opinions? 213.169.163.106 (talk) 20:26, 10 April 2016 (UTC)

April 9

percentages

Can I have an easy explanation for the following...Percentages. Example here.

If I want to calculate say, how much is the pecentage of two numbers i.e what percentage of 400 is against 1400 I get 28.5%. I did this by dividing 400/1400 and moving the decimal places two positions forward.

Ok that's fine. But what if I wanted to know what 40 percent of 6000 was. I'm missing one number already as per example (400) so I've got nothing to start with against the 6000. So it's x/6000.

Am I making this too complicated? looking for a really simple answer to this one. — Preceding unsigned comment added by 82.35.190.215 (talk) 11:12, 9 April 2016 (UTC)

1% of an amount is just 1 hundredth of that amount. So 40% of 6000 is 6000/100 (to get 1%), times 40; 60 x 40 = 2400. Rojomoke (talk) 11:19, 9 April 2016 (UTC)

just remember that percent written in decimal form is, for example 0.40 (0.40=40%)...and percent OF something is multiplication (of=multiplication)...so 0.40 X 6000 = 2400 is a translation of "40% of 6000 is 2400."68.48.241.158 (talk) 12:53, 9 April 2016 (UTC)

just thinking (if you're still around): the general equation you're referencing is: percent/100=part/whole...and then just plug in what numbers you have, solve with "cross multiplication"...68.48.241.158 (talk) 13:24, 10 April 2016 (UTC)

And, if you are using a basic calculator, the "%" button usually is like typing in "/ 100 =". So, you would type in "6000 × 40 %" to get your answer. StuRat (talk) 14:02, 10 April 2016 (UTC)

April 10

geometry problem

This diagram has 8 sections, indicated by the 7 black diagonal lines. I know the total area of the overall rectangle because I am able to measure it with a ruler in the horizontal and in the vertical. (It measures 66 cm vertically and 49 cm horizontally.) I want the area of each of the 8 sections to be equal. I want the black lines to be parallel to each other. How do I calculate what the spacing between the parallel lines should be? Bus stop (talk) 18:49, 10 April 2016 (UTC)

A 66cm x 49cm rectangle split into 8 regions of equal area by 7 diagonal lines

@Bus stop: Does it matter what angle the lines are at? —  crh ₂₃  (Talk) 19:20, 10 April 2016 (UTC)

In the interests of keeping it simple for myself I'd like to start by just trying to do this with the lines intersecting the horizontal and vertical at 45 degree angles. Bus stop (talk) 19:33, 10 April 2016 (UTC)

It may be simpler not to: an easy way is to make it such that the first line comes from the corner to 66/4 cm up the side (giving the triangle the correct area), do the same in the opposite corner and then use the remaining lines spaced apart by 66/8 cm (vertical spacing, not perpendicular) to form parallelograms of the correct size. I'll add a picture if I can draw one. —  crh ₂₃  (Talk) 19:37, 10 April 2016 (UTC)

@Bus stop: As promised, I have drawn a picture and inserted it here. This is a solution to the problem if the lines can be at any angle: this solution produces round lengths, rather than angles. —  crh ₂₃  (Talk) 20:00, 10 April 2016 (UTC)

Crh23—that is very interesting. You have come up with a simple solution to my problem. It is something I had not thought of. I like a lot of things about it from a design perspective. This is for a painting. That is a nice diagram. Thank you.

Note that you could also put the triangles along the long sides of the rectangle, although the angle would be different there. StuRat (talk) 21:41, 10 April 2016 (UTC)

I think I know what you mean, StuRat. Then the parallelograms would all occur along the short side (the horizontal side) of the rectangle. Interesting, and it may make a good design. But I had in mind a 45 degree angle of intersection between diagonal lines and both the horizontal and the vertical. That would also involve varying distances between the parallel lines. Bus stop (talk) 22:41, 10 April 2016 (UTC)

You can also make all the lines parallel to a diagonal. If the rectangle is (0, 0), (a, 0), (0, b), (a, b) then make lines though (a/2, 0) and (0, b/2), (√2a/2, 0) and (0, √2b/2), (√3a/2, 0) and (0, √3b/2), (a, 0) and (0, b), (a, (2-√3)b/2) and ((2-√3)a/2, b), (a, (2-√2)b/2) and ((2-√2)a/2, b), (a, b/2) and (a/2, b). If the angle is given then the problem is messy but doable. You have to integrate a trapezoidal shaped function, but it's linear on each piece if you divide up the domain into three pieces. The integrals on each piece are easy; quadratic functions. You then need to find where it crosses each of the lines kab/8, which means solving some quadratic equations. --RDBury (talk) 22:36, 10 April 2016 (UTC)

Hi RDBury—thanks for the effort you've expended. After researching the word diagonal I realize I was using it incorrectly. I don't mean diagonal. What I mean are lines that intersect with the vertical and the horizontal at 45 degree angles. I'm sorry for the confusion I introduced. Bus stop (talk) 01:28, 11 April 2016 (UTC)

Each of six parallel stripes covers 1/8 of the rectangle area, so the vertical distance v between the lines should be equal 1/8 of the rectangle's height H. Consequently the height of each triangle is 2/8 H. The angle between the slant lines and the rectangle's base is $\arctan \left(\left({\tfrac {2}{8}}H\right):W\right)=\arctan(H:(4W))$. Its cosine is ${\frac {4W}{H^{2}+(4W)^{2}}}$ and the distance (slant) between lines is v times the cosine:
      ${\frac {2}{8}}H\cdot {\frac {4W}{H^{2}+(4W)^{2}}}={\frac {WH}{H^{2}+(4W)^{2}}}$
--CiaPan (talk) 06:27, 11 April 2016 (UTC)

Actually the problem isn't as messy as I thought when the angle is given. Restate and generalize the problem as this: Given a rectangle height H and width W and a number x between 0 and 1, divide the rectangle into two sections of area xHW and (1-x)HX by a line with slope m≠0, so that the xHW area is below and the (1-x)HW area is above. We can take |m|=1 (i.e. 45°) by rescaling the vertical axis. The we can take m=-1 (upper left to lower right) by reflecting through a vertical line. Also assume H≥W, otherwise reflect through a diagonal line. There three possibilities; either the lower left region is a triangle, or the upper right region is a triangle, or both regions are trapezoids (=trapezia outside North America, see a thread from a few weeks ago). In the first case you want an isosceles right triangle of area xHW, and this would have sides √(2xHW) and would work as long as √(2xHW)≤W, which is equivalent to x≤W/2H. For the second case you want an isosceles right triangle of are (1-x)HW and this would have sides √(2(1-x)HW) and would work for x≥1-W/2H. In the third case, the height of the lower trapezoid, meaning the distance between the parallels, is W. Also the difference between the lengths of the parallel sides is W. If u and v are these lengths then you get v-u=W and W(u+v)/2=xHW, which solves to u=xH-W/2, v=xH+W/2 and this works for W/2H < x < 1-W/2H.

For this particular problem, W = 49, H = 66, and x has values 1/8, 2/8, ... 7/8. W/2H is between 2/8 and 3/8 so the first two cuts go through the left and bottom sides of the rectangle. The first section is a right triangle with sides √808.5 ≈ 28.43. The second cut would go through the left and bottom sides at distance √1617 ≈ 40.21 from the bottom left corner of the rectangle. The third cut would go through the vertical sides at distances .25 and 49.25 from the bottom. The fourth cut would go through the vertical sides at distances 8.5 and 57.5 from the bottom. The remaining cuts are symmetrical with the first three. The result is backwards from the picture but I'm assuming that doesn't matter. --RDBury (talk) 10:41, 11 April 2016 (UTC)

April 11

Calculating mean from stddev and deviation

Suppose that I know that a distribution is normal. I know the standard deviation is 16. I know that 88.1% percent of values is <140. I want to calculate the mean. Is there an easy way to do this or does it solving the population density for the mean after filling in X and standard deviation? 199.15.144.250 (talk) 18:09, 11 April 2016 (UTC)