A blog from a couple weeks ago has bothered me. If we can learn from our mistakes, then I have an opportunity here.
I thought I would try finding the area of Colorado. It is a rectangle. That was mistake #1. It is bordered by lines of longitude of 102W and 109W. That is a difference of 7 degrees. but distances between lines of longitude don't stay the same. The distance between them narrows as you move from the equator going toward the poles where they all meet. So Colorado is really more of a trapezoid. And there we have mistake #2. It really isn't a trapezoid since it is on the surface of a sphere. So there are non-Euclidean aspects to deal with. I won't be dealing with that as I'm having enough problems with this. I'll stick with Euclid.
In mistakes #3 and counting, I found that it either can't be done, or I can't figure out how to get those horizontal boundary distances. I resorted to a website where you can plug in latitudes and longitudes and it will compute the distance using something called the haversine formula.
Long story short, doing so told me the south boundary is 386.2 miles and the north boundary is 363.9 miles. A couple weeks ago I had correctly computed the height as lines of latitude are parallel and thus stay the same distance apart. That value was 276.4 miles. Using the formula for area of a trapezoid, I got 103,056 square miles. The internet says it is 103,718 square miles. I'm calling that a win.
I'm not sure how they actually figure the areas of states. Maybe its just an estimate. Online I found at least three different values for the area of the state on sites that seemed to be fairly credible. So, my guess is as good as theirs.
Monday, November 30, 2015
Wednesday, November 25, 2015
Amount of Daylight
We come to the time of the year when there is the minimum amount of daylight. That is depressing, but it's balanced by the fact that it is the holiday season. A trig equation could be written that would show the amount of daylight for each day of the year. It can be done without a whole lot of information.
Sounds like fun. Here we go.
We need to come up with A, k, and c for the equation y = Asin(kx+c).
Some basic information is that there are 365 days in a year. The least amount of daylight is on the first day of Winter - around December 21. The most is around June 21. It will be even amounts on the equinox dates - March 21 and September 21.
With this info, we can examine amplitude, period, and phase shift.
Amplitude - Let's assume we get three extra hours on the first day of summer and three fewer hours on the first day of winter. So A = 3. That was easy.
Period - Since Period = (2pi)/k and the daylight cycle is 365 days, so 365 = (2pi)/k. Therefore k = (2pi)/365 = 0.0172
Phase Shift - It would have been convenient if the spring equinox fell on January first. There would have been no phase shift. Instead, it falls about 80 days later. Phase shift = -c/k, So 80 = -c/0.0172. We get c = -1.376.
Our equation can be written y = 3sin(0.0172x-1.376). The x stands for the number day of the year, such as for January 12, x = 12. For February 1, x = 32. The y stands for the amount of extra or less sunlight. I tried it out for a few values and it seems to work. On December 21 we have three fewer hours. On January 1, we have 2.94 fewer hours - a slight improvement.
We could change the formula so it could stand for the full amount of daylight by tacking on a +12 to the end: y = 3sin(0.0172x-1.376)+12
Tuesday, November 17, 2015
Chance of Being Undefeated
They said that now, at the midpoint of the NFL season, there are still three undefeated teams. That is the most there have ever been at this point in the season. What is the chance we have at least one of those teams staying undefeated for the whole season?
There are seven games to go for each team. First we would want to know what the chance of winning a single game would be. I'm going to go with an 80% chance. That sounds about right. They're undefeated at this point, so obviously pretty good, yet they wouldn't be invincible.
The chance of any one team of winning their next seven games is (0.8)^7 = 0.21. Now what is the probability of New England, or Cincinnati, or Carolina remaining unbeaten? The easiest way to deal with a three team "or" problem is to look at it negatively. There is a 1-0.21 = 0.790 chance of a team having at least one loss the rest of the way. The chance all three are defeated is 0.493. So, there must be a 1-0.493 = 0.507 of that not being the case. So chances are at least one team will be undefeated - a 50.7% chance.
It could be done the straight forward way, but its a little more cumbersome.
0.21+0.21+0.21-(0.21)(0.21)-(0.21)(0.21)-(0.21)(0.21)(0.21)+(0.21)(0.21)(0.21) = 0.507 = 50.7%
For full disclosure, since I thought about it yesterday, Cincinnati lost on Monday Night Football. I was just giving them the win. I probably shouldn't have done that.
Now we're down to two teams. There is still a (0.8)^7 = 0.21 chance of any one team winning seven in a row. The chance that either Carolina or New England does it is:
(0.21)+(0.21)-(0.21)(0.21) = 0.347 = 34.7%
If you take exception to my 80% single win assumption, it's easy to substitute another value. This makes a bigger difference in the final probabilities than I would have thought. For example, if you figure the chance of a team of this caliber wins a single game is 85%, the chance one of the remains unbeaten is now 53.9%. If 90%, the chance of having an unbeaten is all the way up to 73.8% .
There are seven games to go for each team. First we would want to know what the chance of winning a single game would be. I'm going to go with an 80% chance. That sounds about right. They're undefeated at this point, so obviously pretty good, yet they wouldn't be invincible.
The chance of any one team of winning their next seven games is (0.8)^7 = 0.21. Now what is the probability of New England, or Cincinnati, or Carolina remaining unbeaten? The easiest way to deal with a three team "or" problem is to look at it negatively. There is a 1-0.21 = 0.790 chance of a team having at least one loss the rest of the way. The chance all three are defeated is 0.493. So, there must be a 1-0.493 = 0.507 of that not being the case. So chances are at least one team will be undefeated - a 50.7% chance.
It could be done the straight forward way, but its a little more cumbersome.
0.21+0.21+0.21-(0.21)(0.21)-(0.21)(0.21)-(0.21)(0.21)(0.21)+(0.21)(0.21)(0.21) = 0.507 = 50.7%
For full disclosure, since I thought about it yesterday, Cincinnati lost on Monday Night Football. I was just giving them the win. I probably shouldn't have done that.
Now we're down to two teams. There is still a (0.8)^7 = 0.21 chance of any one team winning seven in a row. The chance that either Carolina or New England does it is:
(0.21)+(0.21)-(0.21)(0.21) = 0.347 = 34.7%
If you take exception to my 80% single win assumption, it's easy to substitute another value. This makes a bigger difference in the final probabilities than I would have thought. For example, if you figure the chance of a team of this caliber wins a single game is 85%, the chance one of the remains unbeaten is now 53.9%. If 90%, the chance of having an unbeaten is all the way up to 73.8% .
Tuesday, November 10, 2015
Area of Colorado
I thought it might be interesting to try to compute the area of Colorado by using arc lengths - just to see if it comes out right. It doesn't. At least not how I did it. Maybe its of use to someone that can learn from my mistake(s).
One troubling thing is that I looked on the internet and got three different values for the area of the state. I would think in the age of GPS we would have that figured out to a pretty precise amount. The three amounts were separated by 94 square miles. At least that gives me some wiggle room.
I figured I could find both the length and width of the state would be with:
(arc length)/360(2pi(radius of the earth))
The radius of the earth is 3,959 miles. The arc length I figured would be the differences in the latitudes or the longitudes. The height came out great. I got 276.4, and the web says 276. The width wasn't even close: my 483.7 to their 387. Mine was too big. Arc length would include the curvature of the earth. I thought maybe I could use Law of Cosines to get a closer figure.
C^2 = (3959)^2+(3959)^2-2(3959)(3959)Cos(7)
It was interesting that I got 483.4 as compared to an arclength of 483.7. What was not interesting was the web says the length of Colorado is 387.
What I learned:
1. I noticed their published lengths and widths didn't multiply to get their published area. It was 104,091 to 106,812 square miles. So, they must not use length x width to get the area. That makes sense, because that would only work on rectangular states of which there are not many.
2. Why my method didn't work. I knew this but didn't think about it - lines of longitude are not the same distance apart. They are a certain distance apart at the equator and shrink to nothing at the poles. I guess that is why my north-south distance came out accurately but my east-west was way off.
So, I don't know how they find the area of a state or any region for that matter.
There's a good project for you or your students.
One troubling thing is that I looked on the internet and got three different values for the area of the state. I would think in the age of GPS we would have that figured out to a pretty precise amount. The three amounts were separated by 94 square miles. At least that gives me some wiggle room.
I figured I could find both the length and width of the state would be with:
(arc length)/360(2pi(radius of the earth))
The radius of the earth is 3,959 miles. The arc length I figured would be the differences in the latitudes or the longitudes. The height came out great. I got 276.4, and the web says 276. The width wasn't even close: my 483.7 to their 387. Mine was too big. Arc length would include the curvature of the earth. I thought maybe I could use Law of Cosines to get a closer figure.
C^2 = (3959)^2+(3959)^2-2(3959)(3959)Cos(7)
It was interesting that I got 483.4 as compared to an arclength of 483.7. What was not interesting was the web says the length of Colorado is 387.
What I learned:
1. I noticed their published lengths and widths didn't multiply to get their published area. It was 104,091 to 106,812 square miles. So, they must not use length x width to get the area. That makes sense, because that would only work on rectangular states of which there are not many.
2. Why my method didn't work. I knew this but didn't think about it - lines of longitude are not the same distance apart. They are a certain distance apart at the equator and shrink to nothing at the poles. I guess that is why my north-south distance came out accurately but my east-west was way off.
So, I don't know how they find the area of a state or any region for that matter.
There's a good project for you or your students.
Sunday, November 1, 2015
The Law of Large Numbers and MVP's
As you do repeated trials, the mean average of those trials will approach the theoretical mean. Or if you are looking at a sample, as your sample grows, its mean average will get closer to the mean of the entire population.
The law of large numbers is a good title for this. Many have the idea that the law of probability would imply that a .250 hitter that goes 0 for 3 is now "due" for a hit. Flipping a coin 10 times means you'll get 5 heads and 5 tails. For most of us, our own life experience would show statements like these to be incorrect. It would not be weird for a coin being flipped 10 times to have 3 heads and 7 tails. However, we would think something was up if we flipped it 1,000 times and got 300 heads and 700 tails.
Like many math teachers, I would have classes do some coin flipping experiments. Always a fun day. For years I would write down the results and keep a running total. I don't know where that is now. I wish I had kept that. I was up to something like 20,000 flips. It wasn't 50-50, but pretty close. Maybe something like 49.7% to 50.3%.
I'm reminded of this in something I read in "The Signal and the Noise: Why So Many Predictions Fail - But Some Don't" by Nate Silver. It's an interesting book. One example: At what is a major league baseball player at his peak? You can make a pretty good case for it being 27 years of age. To make his case, he looked at 50 MVP award winners. Granted, 50 is possibly not to be considered a "large number", but it's what was used in this case.
"A baseball player...peaks at age twenty-seven. Of the fifty MVP winners between 1985 and 2009, 60 percent were between the ages of twenty-five and twenty-nine, and 20 percent were aged twenty-seven exactly."
This doesn't prove anything for sure, but then again, surveys never do. I would think we would have a better idea as we can look at additional MVP's in coming decades, giving more applicability to the law of large numbers. Also, the results might have been more convincing if they didn't include Barry Bonds steroid-assisted MVP awards in his late thirties.
The law of large numbers is a good title for this. Many have the idea that the law of probability would imply that a .250 hitter that goes 0 for 3 is now "due" for a hit. Flipping a coin 10 times means you'll get 5 heads and 5 tails. For most of us, our own life experience would show statements like these to be incorrect. It would not be weird for a coin being flipped 10 times to have 3 heads and 7 tails. However, we would think something was up if we flipped it 1,000 times and got 300 heads and 700 tails.
Like many math teachers, I would have classes do some coin flipping experiments. Always a fun day. For years I would write down the results and keep a running total. I don't know where that is now. I wish I had kept that. I was up to something like 20,000 flips. It wasn't 50-50, but pretty close. Maybe something like 49.7% to 50.3%.
I'm reminded of this in something I read in "The Signal and the Noise: Why So Many Predictions Fail - But Some Don't" by Nate Silver. It's an interesting book. One example: At what is a major league baseball player at his peak? You can make a pretty good case for it being 27 years of age. To make his case, he looked at 50 MVP award winners. Granted, 50 is possibly not to be considered a "large number", but it's what was used in this case.
"A baseball player...peaks at age twenty-seven. Of the fifty MVP winners between 1985 and 2009, 60 percent were between the ages of twenty-five and twenty-nine, and 20 percent were aged twenty-seven exactly."
This doesn't prove anything for sure, but then again, surveys never do. I would think we would have a better idea as we can look at additional MVP's in coming decades, giving more applicability to the law of large numbers. Also, the results might have been more convincing if they didn't include Barry Bonds steroid-assisted MVP awards in his late thirties.
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