Alex Plank--What the @#$% is this crap?

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Ganondox
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07 Sep 2012, 10:56 pm

unduki wrote:
Maybe she did overreact but the sexist comments that follow are beyond disgusting.

How dare a woman get out of line!! !

Women have every right to overreact to a society that has largely ignored them forever and continues to do so - unless some man needs his ass wiped, right?

Referencing feminazis is not going to help the situation but I suspect helping others isn't the objective of many here.


The thread would have been taken seriously by a lot more people if the reaction wasn't so over the top. The issue was valid, but the OP makes people think it's just a feminazi whining over nothing.


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07 Sep 2012, 11:24 pm

alex wrote:
He is often used as a gender neutral pronoun. I'm surprised there are people ignorant of a grammatical convention that's been around since the 18th century (or earlier).


I guess I never knew that before. 8O I'm surprised I never learned about it.



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07 Sep 2012, 11:50 pm

alex wrote:
He is often used as a gender neutral pronoun. I'm surprised there are people ignorant of a grammatical convention that's been around since the 18th century (or earlier).


Because nothing from the 18th century is sexist.

This convention is not acceptable today and if you choose not to change it you are choosing to alienate a good number of female readers.

Quote:
The reason I only included the statistic for boys is due to the fact that I believe cases of autism in girls are seriously underrepresented (only 1 in 252 are diagnosed) due to a lot of different factors, so the girls number is misleading, and so is the overall number. I think there are just as many girls with autism, they're just diagnosed less.


Again, it might not be your intent, but this further marginalizes females with AS.

I'm really disappointed by this response. Accusing us of ignorance, really?



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08 Sep 2012, 12:37 am

Dirtdigger wrote:
LostInSpace wrote:
Dirtdigger wrote:
Delphiki wrote:
Dirtdigger wrote:

But if a person does the math a lot more boys have autism than girls.

Autism in girls:
1 in 34

Autism in boys
1 in 54

Do you know math?


What a stupid question. Does 34 and 54 adds up to 88? And is the 1 in each group included? So, lets not get smart and try to understand what's going on here.


Yeah... that's not how averages work. This is how you compute an average of two numbers:

(Number1 + Number2) / 2

If 1/54 boys had AS and 1/34 girls had it, then the average of all children would be between 1 in 54 and 1 in 34, hardly 1 in 88. By my calculation (assuming equal numbers of boys and girls), 11/459 children would have it, or slightly more than 1 in 42. Also, it would be much more common in girls if that were the case (1/34 is more common than 1/54).


I didn't catch that in the article. So you are right. Someone else made a booboo too besides me who simply didn't catch the discrepency. So I divided 2 into 88 since there is one from the girl group and one from the boy group and come up with 44 if there were equal numbers of boys and girls.




NONONONONONO! Bad Dirtdigger! No math for you! If 1/88 people had had autism and the rates are the same than 1/88 boys have autism and 1/88 girls.

Now it's time to do some REAL math. Frequency is essentially the same as probability. The rest is between 0 and 1, 0 means no one and 1 means everyone. 1 in 5 is 1/5, if the top number is larger compared to another ration with the same bottom it's more frequent, if the bottom is larger it's less fequent. To calculate the frequency in a particular group you need to divide those in the group with the condition by the total amount of people in the group. Frequency of people with autism (FP[A]) is People with autism (P[A]) divided by everyone (P), so FP[A] = P[A]/P. Likewise, for boys FB[A] = B[A]/B, and for girls FG[A] = G[A]/G. For the Sake of the problem everyone is either a boy or a girl, so P = B + G and P[A] = B[A] + G[A]. Using basic algebra by multiplying FP[A] by P you get P[A].

For our first problem, let's assume FP[A] = 1/88 and FB[A] = FG[A]. Let's call that number x. Using math we can get that P[A] = B[A] + G[A] = xB + xG = x(B + G) = xP, x = xP/P = P[A]/P = FP[A] = 1/88. Thus 1 in 88 boys would have autism, and 1 in 88 girls would have autism.

Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


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08 Sep 2012, 12:40 am

Nonperson wrote:
alex wrote:
He is often used as a gender neutral pronoun. I'm surprised there are people ignorant of a grammatical convention that's been around since the 18th century (or earlier).


Because nothing from the 18th century is sexist.

This convention is not acceptable today and if you choose not to change it you are choosing to alienate a good number of female readers.


It is acceptable, you are just being ridiculous. Be thankful that English lacks true grammatical gender. This is why feminists aren't being taken seriously.


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08 Sep 2012, 2:01 am

It just keeps getting better, doesn't it?



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08 Sep 2012, 6:22 am

Ganondox wrote:
Dirtdigger wrote:
I think Alex should have been a bit more thorough. I found this on RxList. There are other websites that basically is saying the same things.

Alex probably should have said something like this.

One in every 88 U.S. children -- and one in 54 boys -- has autism, the CDC now estimates.

But if a person does the math a lot more boys have autism than girls.

Autism in girls:
1 in 34

Autism in boys
1 in 54

But, rather going ballistic, I do an Internet search.

1 in 88 children has Autism

I look for the number of Autism children go up as NTs and the mental health professional becomes more familiar with it.


I agree with the first part. In the second part the math is completely wrong. You interpreted "1 in 54" as "54 of the 88" and miscalculated the math from there.


Don't blame me. Blame the article. This article didn't even get it right when they said 1 in 88 and than they turn around and say 1 in 54 boys. I only subtracted 54 from 88 to get the 34 girls. So one from each group has Autism. So get your priorities straight or did you even bother to read the article?



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08 Sep 2012, 6:33 am

Ganondox wrote:
Dirtdigger wrote:
LostInSpace wrote:
Dirtdigger wrote:
Delphiki wrote:
Dirtdigger wrote:

But if a person does the math a lot more boys have autism than girls.

Autism in girls:
1 in 34

Autism in boys
1 in 54

Do you know math?


What a stupid question. Does 34 and 54 adds up to 88? And is the 1 in each group included? So, lets not get smart and try to understand what's going on here.


Yeah... that's not how averages work. This is how you compute an average of two numbers:

(Number1 + Number2) / 2

If 1/54 boys had AS and 1/34 girls had it, then the average of all children would be between 1 in 54 and 1 in 34, hardly 1 in 88. By my calculation (assuming equal numbers of boys and girls), 11/459 children would have it, or slightly more than 1 in 42. Also, it would be much more common in girls if that were the case (1/34 is more common than 1/54).


I didn't catch that in the article. So you are right. Someone else made a booboo too besides me who simply didn't catch the discrepency. So I divided 2 into 88 since there is one from the girl group and one from the boy group and come up with 44 if there were equal numbers of boys and girls.




NONONONONONO! Bad Dirtdigger! No math for you! If 1/88 people had had autism and the rates are the same than 1/88 boys have autism and 1/88 girls.

Now it's time to do some REAL math. Frequency is essentially the same as probability. The rest is between 0 and 1, 0 means no one and 1 means everyone. 1 in 5 is 1/5, if the top number is larger compared to another ration with the same bottom it's more frequent, if the bottom is larger it's less fequent. To calculate the frequency in a particular group you need to divide those in the group with the condition by the total amount of people in the group. Frequency of people with autism (FP[A]) is People with autism (P[A]) divided by everyone (P), so FP[A] = P[A]/P. Likewise, for boys FB[A] = B[A]/B, and for girls FG[A] = G[A]/G. For the Sake of the problem everyone is either a boy or a girl, so P = B + G and P[A] = B[A] + G[A]. Using basic algebra by multiplying FP[A] by P you get P[A].

For our first problem, let's assume FP[A] = 1/88 and FB[A] = FG[A]. Let's call that number x. Using math we can get that P[A] = B[A] + G[A] = xB + xG = x(B + G) = xP, x = xP/P = P[A]/P = FP[A] = 1/88. Thus 1 in 88 boys would have autism, and 1 in 88 girls would have autism.

Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Another fool that is quick to judge and lack the understanding who haven't read all of my posts including the article. I'm not taking the blame for this crap in this article. And knock off the math crap too. I know I'm not good in math so quit throwing it up in my face!. So of you people who pride yourselves in going in for the kill when there is a weakness shows just how unsympathic, with no empathy what so ever you are. I'm only generalizing as if there were 44 boys and 44 girls.

I MADE D'S AND F'S IN MATH. ARE YOU HAPPY NOW?



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08 Sep 2012, 8:02 am

Ganondox wrote:
Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Thank you. I made a very stupid error when I did my calculations, which is pretty much typical for me when trying to do math. But I am happy to see that I understood the principal of what I was supposed to do.


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08 Sep 2012, 8:14 am

InThisTogether wrote:
Ganondox wrote:
Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Thank you. I made a very stupid error when I did my calculations, which is pretty much typical for me when trying to do math. But I am happy to see that I understood the principal of what I was supposed to do.


If this article gave us the right information in the first place there wouldn't be so much confusion and so many heated dabates to the point where some are picking on others once they uncover their weaknesses.



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08 Sep 2012, 8:24 am

Dirtdigger wrote:
Ganondox wrote:
Dirtdigger wrote:
LostInSpace wrote:
Dirtdigger wrote:
Delphiki wrote:
Dirtdigger wrote:

But if a person does the math a lot more boys have autism than girls.

Autism in girls:
1 in 34

Autism in boys
1 in 54

Do you know math?


What a stupid question. Does 34 and 54 adds up to 88? And is the 1 in each group included? So, lets not get smart and try to understand what's going on here.


Yeah... that's not how averages work. This is how you compute an average of two numbers:

(Number1 + Number2) / 2

If 1/54 boys had AS and 1/34 girls had it, then the average of all children would be between 1 in 54 and 1 in 34, hardly 1 in 88. By my calculation (assuming equal numbers of boys and girls), 11/459 children would have it, or slightly more than 1 in 42. Also, it would be much more common in girls if that were the case (1/34 is more common than 1/54).


I didn't catch that in the article. So you are right. Someone else made a booboo too besides me who simply didn't catch the discrepency. So I divided 2 into 88 since there is one from the girl group and one from the boy group and come up with 44 if there were equal numbers of boys and girls.




NONONONONONO! Bad Dirtdigger! No math for you! If 1/88 people had had autism and the rates are the same than 1/88 boys have autism and 1/88 girls.

Now it's time to do some REAL math. Frequency is essentially the same as probability. The rest is between 0 and 1, 0 means no one and 1 means everyone. 1 in 5 is 1/5, if the top number is larger compared to another ration with the same bottom it's more frequent, if the bottom is larger it's less fequent. To calculate the frequency in a particular group you need to divide those in the group with the condition by the total amount of people in the group. Frequency of people with autism (FP[A]) is People with autism (P[A]) divided by everyone (P), so FP[A] = P[A]/P. Likewise, for boys FB[A] = B[A]/B, and for girls FG[A] = G[A]/G. For the Sake of the problem everyone is either a boy or a girl, so P = B + G and P[A] = B[A] + G[A]. Using basic algebra by multiplying FP[A] by P you get P[A].

For our first problem, let's assume FP[A] = 1/88 and FB[A] = FG[A]. Let's call that number x. Using math we can get that P[A] = B[A] + G[A] = xB + xG = x(B + G) = xP, x = xP/P = P[A]/P = FP[A] = 1/88. Thus 1 in 88 boys would have autism, and 1 in 88 girls would have autism.

Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Another fool that is quick to judge and lack the understanding who haven't read all of my posts including the article. I'm not taking the blame for this crap in this article. And knock off the math crap too. I know I'm not good in math so quit throwing it up in my face!. So of you people who pride yourselves in going in for the kill when there is a weakness shows just how unsympathic, with no empathy what so ever you are. I'm only generalizing as if there were 44 boys and 44 girls.

I MADE D'S AND F'S IN MATH. ARE YOU HAPPY NOW?


The article has absolutely nothing to do with how to calculate the figures. If you were so bad at math then why did you go ahead and be so arrogant with proclaiming your mathematical abilities and insult other people when they point out your math makes no sense? The statistic isn't worded badly, that's how they always word things and it really makes perfect sense.


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08 Sep 2012, 8:27 am

Dirtdigger wrote:
InThisTogether wrote:
Ganondox wrote:
Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Thank you. I made a very stupid error when I did my calculations, which is pretty much typical for me when trying to do math. But I am happy to see that I understood the principal of what I was supposed to do.


If this article gave us the right information in the first place there wouldn't be so much confusion and so many heated dabates to the point where some are picking on others once they uncover their weaknesses.


Although the post was directed at you, I doubt it was a personal attack on you. (I am aware that sentence makes no sense, btw). I think, rather, it was a reaction to not following the "rules" of math, and the reaction would have been the same no matter who posted the error. I think many people here would not have been able to explain how to get the right number, hence the reason no one did it until 6 pages in. The fact that you had difficulty in math clearly explains why you made your error. I do not mean that in a derogatory way at all. The math required to do it is not "straightforward," so of course someone who doesn't have math as a strength would not do it correctly. Nothing to feel bad about.

I would imagine if he would have seen my straightforward request for help before he saw your error, he probably would have responded to that. But if he is like me, when he saw the error, he immediately wanted to correct it, not because of who made the error, and not to pick anyone apart, but because the error was made.


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08 Sep 2012, 8:41 am

Ganondox wrote:
Dirtdigger wrote:
Ganondox wrote:
Dirtdigger wrote:
LostInSpace wrote:
Dirtdigger wrote:
Delphiki wrote:
Dirtdigger wrote:

But if a person does the math a lot more boys have autism than girls.

Autism in girls:
1 in 34

Autism in boys
1 in 54

Do you know math?


What a stupid question. Does 34 and 54 adds up to 88? And is the 1 in each group included? So, lets not get smart and try to understand what's going on here.


Yeah... that's not how averages work. This is how you compute an average of two numbers:

(Number1 + Number2) / 2

If 1/54 boys had AS and 1/34 girls had it, then the average of all children would be between 1 in 54 and 1 in 34, hardly 1 in 88. By my calculation (assuming equal numbers of boys and girls), 11/459 children would have it, or slightly more than 1 in 42. Also, it would be much more common in girls if that were the case (1/34 is more common than 1/54).


I didn't catch that in the article. So you are right. Someone else made a booboo too besides me who simply didn't catch the discrepency. So I divided 2 into 88 since there is one from the girl group and one from the boy group and come up with 44 if there were equal numbers of boys and girls.




NONONONONONO! Bad Dirtdigger! No math for you! If 1/88 people had had autism and the rates are the same than 1/88 boys have autism and 1/88 girls.

Now it's time to do some REAL math. Frequency is essentially the same as probability. The rest is between 0 and 1, 0 means no one and 1 means everyone. 1 in 5 is 1/5, if the top number is larger compared to another ration with the same bottom it's more frequent, if the bottom is larger it's less fequent. To calculate the frequency in a particular group you need to divide those in the group with the condition by the total amount of people in the group. Frequency of people with autism (FP[A]) is People with autism (P[A]) divided by everyone (P), so FP[A] = P[A]/P. Likewise, for boys FB[A] = B[A]/B, and for girls FG[A] = G[A]/G. For the Sake of the problem everyone is either a boy or a girl, so P = B + G and P[A] = B[A] + G[A]. Using basic algebra by multiplying FP[A] by P you get P[A].

For our first problem, let's assume FP[A] = 1/88 and FB[A] = FG[A]. Let's call that number x. Using math we can get that P[A] = B[A] + G[A] = xB + xG = x(B + G) = xP, x = xP/P = P[A]/P = FP[A] = 1/88. Thus 1 in 88 boys would have autism, and 1 in 88 girls would have autism.

Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Another fool that is quick to judge and lack the understanding who haven't read all of my posts including the article. I'm not taking the blame for this crap in this article. And knock off the math crap too. I know I'm not good in math so quit throwing it up in my face!. So of you people who pride yourselves in going in for the kill when there is a weakness shows just how unsympathic, with no empathy what so ever you are. I'm only generalizing as if there were 44 boys and 44 girls.

I MADE D'S AND F'S IN MATH. ARE YOU HAPPY NOW?


The article has absolutely nothing to do with how to calculate the figures. If you were so bad at math then why did you go ahead and be so arrogant with proclaiming your mathematical abilities and insult other people when they point out your math makes no sense? The statistic isn't worded badly, that's how they always word things and it really makes perfect sense.


And why are you being such an ass about my weakness when you could have approach me in a more intelligent way. DID I SAY MORE INTELLIGENT? Like I said if you read all of my posts I admitted that I made a mistake. What more do you want? So shut up and crawl back in your hole. And why didn't you see that I was only trying to simplify things with the 44 boys and 44 girls though it isn't perfect. And I gaurantee you aren't perfect by any stretch of the imagination. I can see just how unsympathetic some Aspies are toward others. You are the one that started in on me, just remember that because you didn't have any brains to deal with the situation properly. I hope this thread gets locked.

WOULDN'T YOU KNOW. A 16 YEAR OLD PUNK THAT EXPLAINS WHY HE IS BEING OBNOXIOUS AND ABOVE EVERYONE ELSE.



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08 Sep 2012, 8:49 am

InThisTogether wrote:
Dirtdigger wrote:
InThisTogether wrote:
Ganondox wrote:
Now let's solve for FG[A] assuming FB[A] = 1/55 (the math is easier with 1/55 than 1/54) and FP[A] = 1/88. Now, to calculate the frequency of being both a girl and having autism (FP[G&A]) we use G[A]/P. G[A] is the same as G&A which is the same A&G which is the same as A[G]. To get G[A]/P we just need (P[A]/P) - (B[A]/P], which is FP[A] - FP[B&A]. FP[B&A] = FB[A] * (B/P), and FG[A] = FP[G&A]/(G/P) = (FP[A] - FP[B&A])/(G/P) = [(P/88) - P(FB[A]*[B/P])]/G = ([P/88] - [B/55])/G = (P/88G) - (B/55G). If we assume B = G then P = 2G so FG[A] = (2/88) - (1/55) = (1/44) - (1/55) = (1/[4 * 11]) - (1/[5 * 11]) = (5/[ 4 * 5 * 11]) - (4/[ 4 * 5 * 11]) = (5 - 4)/220 = 1/220. Thus approximately 1 in 220 girls has autism. I believe that actual number is far higher, but whatever.


Thank you. I made a very stupid error when I did my calculations, which is pretty much typical for me when trying to do math. But I am happy to see that I understood the principal of what I was supposed to do.


If this article gave us the right information in the first place there wouldn't be so much confusion and so many heated dabates to the point where some are picking on others once they uncover their weaknesses.


Although the post was directed at you, I doubt it was a personal attack on you. (I am aware that sentence makes no sense, btw). I think, rather, it was a reaction to not following the "rules" of math, and the reaction would have been the same no matter who posted the error. I think many people here would not have been able to explain how to get the right number, hence the reason no one did it until 6 pages in. The fact that you had difficulty in math clearly explains why you made your error. I do not mean that in a derogatory way at all. The math required to do it is not "straightforward," so of course someone who doesn't have math as a strength would not do it correctly. Nothing to feel bad about.

I would imagine if he would have seen my straightforward request for help before he saw your error, he probably would have responded to that. But if he is like me, when he saw the error, he immediately wanted to correct it, not because of who made the error, and not to pick anyone apart, but because the error was made.


Thank you. You have proven that things can be discussed in a very intelligent way. You said nothing out of line what so ever and yes I agree it could have been anyone that posted and made a mistake. But regardless there is a way of telling a person in a nice way like you did without coming across as picking on another.



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08 Sep 2012, 8:51 am

I hope the OP realizes that Alex designed this site.


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