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Zafaron Uriuc
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| | Subject: Herp Derp Derp Thu May 27, 2010 4:14 pm | |
| Alright, people at my school are not very bright. They decided to block gmail and google docs for some stupid reason. And that is the main way I send information from my school to home and back. Since I can't use that now, I guess I will have to temporarily host my essays and stuff on the site. This topic will be what I use for it. Since it will mostly be filled with half finished essays, nobody needs to read past this unless they want to. Anyways - Spoiler:
Ask Marilyn Question
By Derp Derp
Introduction
Sometimes, determining probabilities of events occurring is not as simple as it seems. In 1997, Marilyn Vos Savant, the author of the “Ask Marilyn” column of Parade magazine was asked if the chances of two events occurring were the same for two different situations. The question was referring to the probabilities of two different people having two male children. One woman has two children with one for sure being a boy. The other family consists of a man who has an older child that is a boy. It is assumed that the probability of having a boy or a girl is equal. Most people that look at the problem would say that the chance of having two boys is the same for both families, since it is assured that each will have either one boy or two. However, they are still different situations. While the number of male children possible is the same for both families, there are different numbers of outcomes for each family. By using random digit tables and calculator programs, it is possible to simulate the different samples that could be found for a random family and calculate the probability of successfully having to male children.
Analysis
One of the simplest, but most time consuming ways to simulate taking various samples of two children is to use a table of random numbers. The first step was to randomly choose one of the lines of the table to begin with. After choosing a line, two numbers were taken at a time from the line and examined. For deciding the gender of the woman’s two children, having a pair of numbers that are both even designates that the woman has two male children. If the number from the table has one even digit and one odd digit, with order not mattering, then it represented that the woman has one male child and one female child. If both numbers were odd, then it represented that the woman has two female children. Since the original question states that at least one of the children is male, any number composed of two odd digits was skipped over and ignored. These steps were repeated until one hundred numbers that had at least one even digit were taken from the line. Using Line 115, for this simulation it was determined that twenty eight families had two boys while the other seventy-eight only had one. This demonstrated that for this simulation twenty-eight percent of the families had two boys.
The same process was used to determine the probability of having two boys when the older child is male. For this simulation, a single number was taken from the table of random digits with an even number representing having a male child and an odd number representing a female child. Since the probability of having a boy or girl is equal, it was assumed that around fifty percent of the one hundred samples would have a male child. When the above mentioned procedure was used starting at line 125, forty-three percent of the one hundred families simulated had two male children. While this result was different than the theoretical probability of fifty percent, this is because samples vary around the theoretical mean.
The same steps can be performed using a calculator rather than a random number table. To simulate the woman's situation it is necessary to set up the random generator to generate one hundred fifty numbers between one and four. The number one would represent that the family has two female children and is skipped. Having a two or three means that the family has one male child and one female child with only the order differing between the two number. Having a four represents successfully having two male children. While only one hundred results are needed, the random generator is set up to generate one hundred fifty number due to having to skip over any ones.
For the man's family, the random number generator on the calculator would be set up to generate one hundred results of either one or two. The number of twos then shows how many of the simulated families had two male children. By dividing this by one hundred, the percent of families with two male children can be found.
Using the first program located in the Appendix, the calculator can also be used to determine the outcomes for the woman's family in each trial. The rand function will generate some decimal number between zero and one. The program is then set up to give a result for the two children. In this case, the letter G represents a girl and the B represents a boy. If the number is less then 0.25, the program will show a result of GG. If the number is less than 0.5 but not less than 0.25 the program will display GB. For a number less then 0.75 but not less than the above numbers, then it will display a BG. If none of the above is generated, then it displays the desired result of BB.
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| | | | Zafaron Uriuc
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| | Subject: Re: Herp Derp Derp Fri May 28, 2010 3:04 pm | |
| - Spoiler:
Ask Marilyn Question
By Derp Derp
Introduction
Sometimes, determining probabilities of events occurring is not as simple as it seems. In 1997, Marilyn Vos Savant, the author of the “Ask Marilyn” column of Parade magazine was asked if the chances of two events occurring were the same for two different situations. The question was referring to the probabilities of two different people having two male children. One woman has two children with one for sure being a boy. The other family consists of a man who has an older child that is a boy. It is assumed that the probability of having a boy or a girl is equal. Most people that look at the problem would say that the chance of having two boys is the same for both families, since it is assured that each will have either one boy or two. However, they are still different situations. While the number of male children possible is the same for both families, there are different numbers of outcomes for each family. By using random digit tables and calculator programs, it is possible to simulate the different samples that could be found for a random family and calculate the probability of successfully having to male children.
Analysis
One of the simplest, but most time consuming ways to simulate taking various samples of two children is to use a table of random numbers. The first step was to randomly choose one of the lines of the table to start from. After choosing a line, two numbers were taken at a time from the line and examined. For deciding the gender of the woman’s two children, having a pair of numbers that are both even designates that the woman has two male children. If the number from the table has one even digit and one odd digit, with order not mattering, then it represented that the woman has one male child and one female child. If both numbers were odd, then it represented that the woman has two female children. Since the original question states that at least one of the children is male, any number composed of two odd digits was skipped over and ignored. These steps were repeated until one hundred numbers that had at least one even digit were taken from the line. Using Line 115, for this simulation it was determined that twenty eight families had two boys while the other seventy-eight only had one. This demonstrated that for this simulation twenty-eight percent of the families had two boys.
The same process was used to determine the probability of having two boys when the older child is male. For this simulation, a single number was taken from the table of random digits with an even number representing having a male child and an odd number representing a female child. Since the probability of having a boy or girl is equal, it was assumed that around fifty percent of the one hundred samples would have a male child. When the above mentioned procedure was used starting at line 125, forty-three percent of the one hundred families simulated had two male children. While this result was different than the theoretical probability of fifty percent, this is because samples vary around the theoretical mean.
The same steps can be performed using a calculator rather than a random number table. To simulate the woman's situation it is necessary to set up the random generator to generate one hundred fifty numbers between one and four. The number one would represent that the family has two female children and is skipped. A two or three means that the family has one male child and one female child with only the order differing between the two numbers. Having a four represents successfully having two male children. While only one hundred results are needed, the random generator is set up to generate one hundred fifty numbers due to having to skip over any ones.
For the man's family, the random number generator on the calculator would be set up to generate one hundred results of either one or two. The number of twos then shows how many of the simulated families had two male children. By dividing this by one hundred, the percent of families with two male children can be found.
Using the first program located in the Appendix, the calculator can also be used to determine the outcomes for the woman's family in each trial. The rand function will generate some decimal number between zero and one. The program is then set up to give a result for the two children. In this case, the letter G represents a girl and the B represents a boy. If the number is less then 0.25, the program will show a result of GG. If the number is less than 0.5 but not less than 0.25 the program will display GB. For a number less then 0.75 but not less than the above numbers, then it will display a BG. If none of the above is generated, then it displays the desired result of BB. Using this program for fifty trials resulted in fifteen BGs, seventeen GBs, and eighteen BBs. Of fifty families simulated, thirty-six percent of them had two boys.
The program “Frequency” can easily determine the proportion of a number of trials that results in having tow male children. The user determines how many trials will be completed when initiating the program by defining the variable N. The program will then choose a random number between zero and one. If the number chosen is less than 0.25, and thus would represent that the family has two girls like in the above program, the program ends and will results. If it is above 0.25 the variable J that starts at zero will increase by one. If the number is greater than 0.75 the variable C also will increase by one. This continues until J equals the same number as the trials specified by N. After the trials end, the number of trials that had two male children, determined by C, is divided by the total number of trials, N, to find the frequency at which the family has two male children.
This procedure was used ten times to determine an average of the various relative frequencies found. The results are located in Table 1 in the Appendix. The average of the relative frequencies was 0.33, very close to the results found from the other types of trials.
Conclusion
All of these trials give very similar answers for the probability of having two male children if one of them is certain to be male. The average of all the different types of trials is 0.32. This result makes perfect sense when looking at the type of outcomes that can be found. While the man’s family has two results, either having the younger child be male of females, the woman’s family actually has three different outcomes. These outcomes are having the older child be male and the younger one female, the older child female and the younger child male, or both children are male. Since there is equal probability of each of these events happening, the probability of any of these three outcomes should be 0.33. The result of 0.32 for the probability of having two male children provides evidence of this claim. So for the original question posed to Marilyn Vos Savant, the chance of having the woman have two boys does not actually equal the chance of the woman having two boys. Instead, there is a 0.33 probability that the woman has two boys while the man has a larger 0.5 probability of having two boys.
Appendix
Program: Girls and Boys
:rand->X
:If X<0.25:Then:Disp “GG”
:Else X<0.5:Then:Disp “GB”
:Else X<0.75:Then:Disp “BG”
:Else:Disp “BB”: End
Program: Frequency
:ClrHome
:Disp “HOW MANY TRIALS”
:Prompt N
:0->C:0->J
:While J<N
:rand->X
:If X<0.25:End
:If X>0.75:C+1-> C
:J+1->J
:END
:Disp “REL.FREQ.=”
:Disp C/N
Table 1
Trial Number 1 2 3 4 5 6 7 8 9 10
REL
FREQ 0.32 0.26 0.33 0.27 0.37 0.39 0.34 0.33 0.34 0.32
Total Average
0.33
| |
| | | | Zafaron Uriuc
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| | Subject: Re: Herp Derp Derp Wed Jun 09, 2010 4:07 pm | |
| Pineapple Problem
Alex Silberman
Introduction
For both consumers and producers, the size of the produce grown is important in determining what price should be set to purchase it. For the Drole Pineapple Company,it is necessary to determine the average size of the pineapples to determine the average size of the pineapples they grow in a timely and low-cost manner. It is inefficient to weigh every single pineapple individually, so it is necessary to take a sample or several samples to determine what the sample average is. While a single sample does give data, samples vary giving different results each time. To balance this out, it is necessary to either take enough samples that the variability of each sample cancels each other out, or increase the number of subjects sampled to bring it closer to the total population.
Once an approximate average has been determined, other information can be garnered from the data. One strategy often used by corporation to increase profits is to display some of the larger fruits to make the shopper think they are getting a good deal for the price. To determine what size pineapples would be bigger than the average but also able to find easily, the sample data can be used to find the probability of certain sizes. From there, the company can choose a number that has a small probability of finding, but not too small as to be too difficult to find.
The sample data can also be used to determine if certain changes to the environment of the produce grown actually significantly increases or decreases the size of the entire batch of food. While any change might make a chance for the total population, due to sample variability some of the changes may be due to chance rather than reacting to external changes. It is necessary to test whether or not the data gathered actually proves significance of external factor or is due to chance alone. Using these different tests and studies, statisticians of the Drole Pineapple Company can determine multiple things about one specific field of pineapples.
Analysis
For the first part of this study, we are dealing with one field of pineapples that has already been tested and sampled to determine the mean and standard deviation. For this specific field, the crop size is approximately normally distributed with a thirty-one ounce mean and a two point five standard deviation. N(31,2.5)
To find the probability of selecting a pineapple that weighs more than 35 ounces, it is necessary to first find the Z-score using the mean weight and the selected weight. By dividing the difference of the sample weight and the mean weight by the standard deviation, it is found that the Z-score is 1.6. Using the standard deviation tables or a calculator, the percentage of the normal distribution curve to the right of the 1.6 standard deviation is found to be 0.0548 or 5.48%. This shows that the probability of finding a pineapple that weighs more than or equal to thirty-five pounds from the field is 5.48%.
(35-31)/2.5 = 1.6 Normalcdf(1.6,1x10^99) = 0.0548
Since the probability of finding one pineapple greater than 35 ounces has been determined, the probability of not finding one of these pineapples can be determined to be 94.52%. Using this, it is possible to calculate what the probability of finding certain combinations of pineapples is. For | |
| | | | Zafaron Uriuc
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| | Subject: Re: Herp Derp Derp Thu Jun 10, 2010 4:13 pm | |
| One of the first things that a person learns at any school are the various rules. Every school has its own specific set of rules and regulation that the students are required to obey. For some schools there are certain rules that you are not allowed to utter while others have certain areas you can't enter. Part of the reason for these rules is to keep peace in the class rooms or protect the students. These rules are perfectly reasonable and should be implemented. However, others are enacted for the sole reason of giving the appearance that the administration is doing something to solve problems, even if they are actually useless and ignored. The most defining part of aohjhjs has been the various rules that shape the lives of the students, especially the rules that are not actually written.
One of the largest problems with aohjhs is that the administration likes to create rules all the time, even if they don't actually enforce them or they are actually hamper the educational experience. One of the greatest examples of weak, unenforced rules is the dress code. throughout the school year the students are subjected to various assemblies and meetings what can and cannot be worn on campus. However, every day it is possible to see large numbers of people ignoring the dress code and not being punished. And once a few people aren't punished, others will follow their example believing to be safe. The electronics policy is also irregularly enforced. Some teachers will take any ipods or cellphones that are seen o heard, while others will just willfully ignore it. Some faculty members will even approve the use of musical devices when working on in-class assignments or studying. When so many teachers do not carry out "the law", it loses its validity. To compensate for the fact that some rurles aren't followed, the administration then makes more rules which often are also not followed since so many others violated previously. This leaves a frustrating glut of useless rules that very few people actually follow.
Unlike the written rules, the unwritten ones are much more easily obeyed. One of the main ones is the push by the teachers and counselors for students to primarily focus on academic classes rather than electives. since there is no limit on the number of AP classes someone can take, each non-AP class will bring down a person's class rank and total GPA.For people who are worried about grades and colleges, this will prevent them from even considering taking an elective. However, taking an elective can be extremely rewarding and help students investigate their likes and dislikes. this can help spark career interests in fields students would normally not even consider. However, instead of being seen as more interesting classes, the electives are instead considered to be populated by 'slackers" who want an easy A. They are often considered to be classes for underachievers, even when they can be difficult creative classes. Most students don't even know about several of the electives offered due to this, and only find out about them when there is a large event like the film festival.
However, there is a rule with even greater dominance than the invisible rule that academics are better than electives. This law is that sports, specifically males sports, are more interesting and gain more recognition than anything academic related. In the newsletters, announcements, and on the front page of the school website, sports reign supreme. If any of the male sports play a good game or beat a rival school, there will be a large amount of fanfare. However, when Academic League manages to win enough games to be ranked third in San Diego county, no attention or support is given. Even when individuals manage to get large honors such as scholarships or advancing in some contest there is absolutely no recognition. While some people may argue that highlighting a single person's achievement in a competitive field such as academics isn't right, it is often done for people who do well in sports which nullifies that argument. Altogether, the school has set it up so that sports get the priority over academics and academics get priority over electives.
These written and unwritten rules have created a claustrophobic environment for the people at this school. At the beginning of our high school career the students are told that as high-schoolers they have the ability to choose their own path within a large expanse of opportunities. However, this is not the truth. While the large expanse exists, there are large barriers created by the rules and regulations that the school sets for its inhabitants. To try to get past these barriers will earn the label of "rule breaker" and "trouble maker", even if it is for self-expression and harms no one. At the same time, in the parts of the expanse we are allowed to navigate, there is already a path carved for us by the school itself. And for the people who step off that path to make their own, there is little to no support to be found. This problem will probably never be fully removed from the system, but luckily most teenagers are stubborn enough to chart their own path regardless of the wills of those around them. However, it is a grave disservice for those who can't. | |
| | | | Zafaron Uriuc
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| | Subject: Re: Herp Derp Derp Fri Jun 11, 2010 3:01 pm | |
| Pineapple Problem
Alex Silberman
Introduction
For both consumers and producers, the size of the produce grown is important in determining what price should be set to purchase it. For the Drole Pineapple Company,it is necessary to determine the average size of the pineapples to determine the average size of the pineapples they grow in a timely and low-cost manner. It is inefficient to weigh every single pineapple individually, so it is necessary to take a sample or several samples to determine what the sample average is. While a single sample does give data, samples vary giving different results each time. To balance this out, it is necessary to either take enough samples that the variability of each sample cancels each other out, or increase the number of subjects sampled to bring it closer to the total population.
Once an approximate average has been determined, other information can be garnered from the data. One strategy often used by corporation to increase profits is to display some of the larger fruits to make the shopper think they are getting a good deal for the price. To determine what size pineapples would be bigger than the average but also able to find easily, the sample data can be used to find the probability of certain sizes. From there, the company can choose a number that has a small probability of finding, but not too small as to be too difficult to find.
The sample data can also be used to determine if certain changes to the environment of the produce grown actually significantly increases or decreases the size of the entire batch of food. While any change might make a chance for the total population, due to sample variability some of the changes may be due to chance rather than reacting to external changes. It is necessary to test whether or not the data gathered actually proves significance of external factor or is due to chance alone. Using these different tests and studies, statisticians of the Drole Pineapple Company can determine multiple things about one specific field of pineapples.
Analysis
1. For the first part of this study, we are dealing with one field of pineapples that has already been tested and sampled to determine the mean and standard deviation. For this specific field, the crop size is approximately normally distributed with a thirty-one ounce mean and a two point five standard deviation. N(31,2.5)
To find the probability of selecting a pineapple that weighs more than 35 ounces, it is necessary to first find the Z-score using the mean weight and the selected weight. By dividing the difference of the sample weight and the mean weight by the standard deviation, it is found that the Z-score is 1.6. Using the standard deviation tables or a calculator, the percentage of the normal distribution curve to the right of the1.6 standard deviation is found to be 0.0548 or 5.48%. This shows that the probability of finding a pineapple that weighs more than or equal to thirty-five pounds from the field is 5.48%.
(35-31)/2.5 = 1.6 Normalcdf(1.6,1x10^99) = 0.0548
2. Since the probability of finding one pineapple greater than thirty-five ounces has been determined, the probability of not finding one of these pineapples can be determined to be 94.52%. Using this, it is possible to calculate what the probability of finding certain combinations of pineapples is. For example, if two pineapples are selected, it is possible to determine the probability for having exactly one of them weigh more than thirty-five ounces. Since it has been determined what the probability for selecting a pineapple of that weight is, 0.0548, it can then be multiplied by the probability that the second does not weigh thirty-five ounces, 0.9452. However, since the heavier pineapple can be either the first or the second pineapple, the two probabilities have to be multiplied by each other, and then by two. This will gives a resulting probability of 0.104.
(2C1)(0.0548)(0.9452) = 0.104
3. A random digit table was used to simulate picking pineapples to find one that weighs over 35 ounces. Since the probability of finding a pineapple that heavy is 0.0548, the digits 00 to 05 on a random digit table are set to signify the proper pineapple. The digits 06-99 on the other hand are set to signify a pineapple that is not greater than thirty-five ounces. The line 127 was chosen at first and two numbers were taken at a time until one of the numbers from 00 to 05 was found. The number of two digit numbers necessary to find a heavy pineapple was recorded and was found to be fifty-six for the first trial. Continuing using the same line, the second trial took only six pairs of numbers to find a proper pineapple, clearly illustrating how samples vary.
By repeating this several times, the average number of tries necessary to find a suitable pineapple was determined. Continuing using the line 130 took 19 times. Restarting from line 120 took 11, 36, and then 34 times. Using line 110 then gave results of 31, 14, and 18. The average of these was found to be 25 trials.
4. Despite the fact that the average trial number necessary is 25, it is also possible to find the approximate probability of having at least one pineapple in a group of three randomly selected pineapples. While it was possible to find the probability of all combinations of at least one pineapple, it was actually much easier to find the probability of not having one pineapple weighing more than thirty-five ounces. Since the probability of not having a pineapple over thirty-five ounces is 0.9452, it was then cubed to find the probability that three pineapples do not weigh over thirty-five ounces. By subtracting this number, 0.844, from one, the probability that at least one pineapple in a group of three weighs thirty-five ounces was found to be 0.155
(0.0548)^3 = 0.844 1-0.844= 0.155
5. When a new irrigation system was used on the crops, it was necessary to determine if the mean weight was actually influenced by it. It was first decided how many samples would need to be taken to find a range of mean weights that would be at least within one ounce of the real mean weight with a 95% confidence level. This means that if many samples were taken, 95% of the samples would have means within one ounce of the actual mean. To determine the number of samples necessary to get this margin of error, it was necessary to multiply the Z-score for a 95% confidence, 1.96, by the SE and have it equal one. The steps necessary to find the variable n are located in the Appendix. The number of samples was found to be 25 samples.
6. A 95% confidence interval was also found for the new fields. Using the Z-score of 1.96, the standard deviation of 2.5 ounces, and the sample size of 50 pineapples, a confidence interval was found to determine if the original mean was encompassed by the interval or if there was a significant difference. By adding and subtracting the margin of error of 1.96 times 2.5 divided by the square root of 50 from the new mean of 31.6, an interval of 30.9 to 32.29 was created. Since this does include the original mean of 31, the confidence interval does not prove sufficient evidence of a change in the mean weight of pineapples.
31.6 +/- (1.96)(2.5)/(50^(1/2)) = 30.9 to 32.29
7. A significance test was also used to test if the change in irrigation is significant. By dividing the difference of the two means by the standard error, it was possible to find a Z-score. Using the calculator standard deviation table or a calculator, the percentage of the curve to the right of the 1.697 Z-score was found to be 4.48%. Since this is less than 5%, the new mean is significant to the 5% level.
However, it cannot be said that the irrigation caused the increase in weight. If the only difference was the change of irrigation, then causation could be determined. However, any other change in the environment or other external forces also contributes to the weight of the pineapples. So while there was a significance change that did not occur due to chance, it may or may not be related to the new irrigation system.
Conclusion
In the end, if a person manages to determine the mean and standard deviation of a series of samples that are normally distributed, there are all kinds of data that can be determined. The probability of selecting different weight sizes of the | |
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