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With 10000 repeats of 1000 samples giving a maximum for the highest month, then giving frequencies for each maximum
of the highest month. My first proper run, after debugging of this, had a result of 114 for the 99.5th percentile,
but it was very close to it being 115 as the 99.5th percentile:
84 frequency 0
85 frequency 0
86 frequency 0
87 frequency 0
88 frequency 5
89 frequency 33
90 frequency 92
91 frequency 173
92 frequency 308
93 frequency 491
94 frequency 589
95 frequency 761
96 frequency 793
97 frequency 875
98 frequency 872
99 frequency 842
100 frequency 725
101 frequency 691
102 frequency 590
103 frequency 463
104 frequency 344
105 frequency 315
106 frequency 268
107 frequency 202
108 frequency 141
109 frequency 118
110 frequency 84
111 frequency 79
112 frequency 48
113 frequency 25
114 frequency 24
115 frequency 11
116 frequency 13
117 frequency 12
118 frequency 3
119 frequency 4
120 frequency 1
121 frequency 1
122 frequency 2
123 frequency 1
126 frequency 1
Median: 99
99.5th percentile: 114
Notice that 49 items are 115 or higher. With only one run of the code 115 could still be argued to be a good answer.
Technically the 99.5th percentile is 114, but it is so close that more data is needed.
EDIT/UPDATE: With an adjustment to the program to make it do 100,000 runs I have decided that 115 is the 99.5 percentile,
but it is not exactly that clear cut even then, with 400 runs of 116 or higher, and 553 runs of 115 or higher.
Assuming that run of 100,000 is typical though it is reasonable to call the 99.5th percentile 115.
So in a two tailed strong result (at the 1% significance level) for this at least 115 is needed in the highest month.
In the one tailed version at the 1% significance level, at least 113 is needed for a strong result according to the 100,000 repeats run.
(That is to say the 99th percentile was 113.)
For a two tailed (97.5th percentile) moderate evidence (5% level significance) result at least 110 was needed.
For a one tailed (95th percentile) moderate evidence (5% level significance) result at least 108 was needed.
The 90th percentile was 105, the median (50th percentile) was 98.
I think he means concur in the sense of to agree with. He agrees with bobbym's answer to your question.
My personal view is this:
Some aspects of geography are scientific such as physical geography.
In fact some physical geography really is probably best thought of as a branch of physics and physics in turn a branch of science.
Population geography has some scientific aspects of it: eg. social science
These are never going to be exact sciences, because for starters they depend upon things like the actual locations
of continents, oceans, the equator, poles, water cycles (physical geography) or
things like political decisions, human decisions, artifical constructs, town and city locations, etc. (population geography).
However there are some interesting theories that geography experts have had that are scientific in nature.
It is interesting that "applied mathematics" is a very similar subject to "physics".
You could argue that the curve between the points is not really justified and that perhaps a straight line between them
is better so you could start with the frequency chart and then draw a triangles and rectangles chart from it by joining
the points (X, Frequency) for X = 3 to X = 8.
X Frequency
3 1
4 3
5 6
6 10
7 15
8 21
The rectangle areas are: 1 + 3 + 6 + 10 + 15 = 35
The triangle areas are: 0.5 * (2 + 3 + 4 + 5 + 6) = 10
For the 90th percentile:
45 * 0.9 = 40.5
The first four rectangles are: 1 + 3 + 6 + 10 = 20
The first four triangles are: 0.5 * (2 + 3 + 4 + 5) = 7
Total of first four = 27
So 13.5 more needed to make up the 90th percentile.
This is out of 18 remaining.
Using a diagram I concluded that 15x + 3x^2 = 13.5
The valid solution is x = 0.778719262151...
Hence by adding 7 to this my perecentile estimate was 7.7787 (to 4 d.p) (assuming I haven't made any mistakes with that)
By a similar argument for the 50th percentile (the median) I am getting 6.683281573
or 6.6833 to 4 d.p.
The discrete integer median is 7 and the discrete integer 90th percentile is 8.
The non integer versions are really approximations based upon a continuous curve that either passes through the
points or gives a good approximation of the distribution. In this case I have made a curve through the points for the
first pair of estimates, and a series of straight lines for the second.
The orignial solution that I gave to the 12 months maximums problem used an approximation using the normal
distribution model without any transformation apart from using an appropriate standard deviation and mean.
It assumed that the distribution was symmetrical, whereas in reality it had a right skew caused by the greater capacity for
maximums higher than the mean compared to a squashed distribution below the mean. I could try transforming the
normal curve and some statistics software you can get nowadays makes this easy to do. I should think that the 99.5th
percentile is either 114 or 115, but it is difficult to be more exact without better software or writing some code myself
to do a better simulation. For the time being I am calling it 114.
(That was the 99.5th percentile estimate of maximums taken from 1000 samples put into 12 month categories randomly.)
With the case of taking 3 items from 8 numbers from one to 8 that is n = 3 and m = 8, and with the example of trying
to calculate the 90th percentile I have worked out this illustration of using a continuous approximation and using the
area under the curve:
A good approximation to the frequencies is:
By the area under a curve definition:
Using the area under the curve formula and by solving the cubic polynomial I am getting p = 7.7803 (to 4 d.p.)
Looks like a sensible answer and close to my rough estimate in the previous post.
By a similar process using the above approximation I am getting 6.6912 (to 4 d.p.) for the median.
I have checked your formula for the mean in the case: n = 3 and m = 8
I worked out the mean of 6.75 using my rather tedious method, and your formula also reassuringly gave 6.75 for this case.
I will therefore assume that the formula is correct and works for the mean.
Now what about median (50th percentile) and other percentiles like 99th percentile (and 99.5, 95, 97.5) ?
The question for that part is harder than I had bargained for because it is not a continuous set that we are
taking the numbers from and we have formulas that are correspondingly discrete so we cannot just integrate
the formula and use an area under the curve argument to define the percentile such as for the median:
median = (point at which area up to that point from lowest range value = 0.5)
With n = 3 and m = 8:
The boring answer is 7 on the basis of ((56 + 1)/ 2) = 28.5 then take the average of the 28th and 29th data item of 56 items.
This is okay for an integer version of the median problem, but some conventions might want a curve fitted then some
transformed normal or another continuous approximation used.
If we use the integer version then anything from about the 64th percentile onwards the answer is just 8.
I was hoping for a continuous analogue solution to give some answer inbetween 7 and 8 for 80th, 90th, 95th percentile etc....
In other words something like for the 90th percentile: (56+1)*0.9 = "51.3 th item" (informally)
Then some clever weighted average is done and ends up with something like 7.73
eg. (27 * 8) + (10 * 7 ) = 286
then 286 / 37 = 7.72972.... = 7.73 to 2 d.p.
based on the 63th to 64th cut off between 7 and 8, and the fact that (90 - 63) = 27 and (100 - 90) = 10
then invert the balance with 27 * 8 and 10 * 7 and divide by the frequency.
However that was just one thought I had about that particular case which might give some strange results in others.
It either needs formalizing. Or maybe it just doesn't give the right statistical answers.
It is difficult enough trying to do that for a simple case. The months case is a lot more difficult because there are
two phases of random generation: First populate 12 values using 1000 samples, then take the maximums, and
then repeat a large number of times (eg. 100, 1000, etc.) and work out a 99th percentile or perhaps the 99.5th percentile.
It would be a struggle to derive a general formula for a percentile for the case n = 3 and m = 8.
That was with a uniform sequence {1,2,3,4,5,6,7,8} and select 3 and take the maximum.
The set with the months is not a uniform set of 12 numbers. They are randomly distributed multinomially from 1000 samples.
I'm not sure whether this is the official answer as such, but remember that the bijection does not have to be unique
it merely has to exist so:
1 - 2
2 - 3
3 - 5
4 - 7
5 - 11
6 - 13
7 - 17
8 - 19
9 - 23
10 - 29
(For each addition of one of the left hand number, the right hand number is the next prime number in numeric order)
Keep on going with the natural numbers from 1 to infinity. Give each of them the next prime in sequence.
If you were to run out of prime numbers then this would contradict the proven theory that there must be
an infinite number of them otherwise you could multiply the finite set of them and add one to generate another.
You cannot run out of the natural numbers either because if you did then just take the highest one and add one
and you have another. Also you must always be able to continue indexing them like this without missing any
out of the list. For any given prime we will be able to give it an index if you had unlimited computing power and every
index has a prime number associated with it according to the numeric ordering of primes.
The fact that the computer would not be powerful enough to do this is irrelevant - in theory it still has
a unique index for each and every prime number, and a unique prime number for every index.
Hence the cardinality is the same as natural numbers, integers and rational numbers.
Thanks Bob.
I have just read an interesting wikipedia article on a related matter concerning cardinality of the reals and other sets:
http://en.wikipedia.org/wiki/Cardinalit … _continuum
It says that Georg Cantor proved the matter in 1874. Interestingly the power set always has higher infinite cardinality
than the original infinite set. Also the power set of the integer set has the same cardinality of the reals I notice.
I did wonder whether you meant the power set of the prime numbers. That would be the same as the power set of
the naturals. That I suppose would be the same as the reals. However Bob's post suggests that it was just the set
of primes in the {7/3} question.
I have never seen the power set property proven. Disappointingly I don't find the wikipedia explanation very clear
at the current time. I suppose if I worked on it a bit it might make more sense. I had previously assumed that the
set of all subsets of a set had equal cardinality to the original set in the infinite case, but this is not true.
I wouldn't like to prove the result for the cardinality of sets of subsets of an infinite set, but yes it sounds like the
wikipedia article is correct. I suppose textbooks still have a use then - because online sources do not always explain
these things very clearly even if the facts are correct.
The cardinality of the number of primes in total is countably infinite. That is they can have a bijection with the
natural numbers. Consider listing them with the natural number as an index reference and the prime as the
item. The cardinality result is then obvious.
However I am not exactly sure at this time as to whether that was the question. The icon has obscured it to some extent,
also I think more words would have been a good idea. Did you mean some property of subsets of primes?
I think I might have just have thought about what is not obvious about it - I presume you haven't done the theorem
concerning cardinality of rationals being the same as the cardinality of integers and natural numbers. In which case
yes intuitively the primes may seem like a smaller set than the natural numbers, but it is known to be not finite,
so how could we have a smaller set? (I don't think we can)
The strange aspect of the Cantor/Schroeder/Bernstein results is/are that the real numbers have higher cardinality,
that is a strange result after the natural/integer/rational result. There are more real numbers than integers to the
extent that you can never list a complete set of them, or something like that - rephrase that more precisely perhaps.
I cannot remember how you prove the result with the reals, but the rational proof draws a diagram involving a sort
of rotation around the 2 dimensional number plane in which the p/q rationals in simplest form are "listed" in a
systematic manner. The proof involving reals being higher in cardinaility involves a method which I have a vague memory
of involving using the fact that there are an infinite number of numbers available after each point in an infinitely accurate
decimal representation. I think it uses a contradiction argument in a generalized listing to prove that some items must
have been missed out and that therefore no such listing can be made.
Compared to that the prime number set cardinality is quite simple which is why I wondered whether there was a result
involving finite cardinality of subsets of the primes constrained somehow, but there is not enough information in the
question to ask such a thing.
Okay. I might leave a few more questions on here later. Bye for now.
A few on volume for you to puzzle over:
(1) What is 500 millilitres converted into centilitres?
500 / 10 = 50 centilitres
(2) What is 500 millilitres converted into litres?
500 / 1000 = 0.5 litres
(3) What is 200 litres in cubic metres?
200 / 1000 = 0.2 cubic metres
(4) What is one pint converted into litres? (Hint: there are 1.7598 pints in one litre. This is using UK conventions not US conventions.)
1 / 1.7598 = 0.5682.... litres
So there are 0.568 litres in a pint to 3 d.p. using the UK imperial pint definition.
(Apparently the US convention is that there are 473 millilitres in a US liquid pint. In UK there are more millilitres than that
in a UK imperial pint. It is the UK version that I was referring to in question 4)
Notice that for instance with question (4): no of litres x 1.7598 = no of pints
no of pints / 1.7598 = no of litres
I was taught to imagine 'dividing both sides of the equation by 1.7598' to get the second. More on that when you do algebra.
(I have improved #1236 a bit let me know if you don't agree with any of it. However I believe it is all correct)
How about converting 86 Fahrenheit into Celcius ?
86 - 32 = 54
54 / 9 = 6
6 x 5 = 30
Answer: 30 C
Also what is 100 Celcius in Fahrenheit ?
100 / 5 = 20
20 x 9 = 180
180 + 32 = 212
Answer: 212 F
What is 104 Fahrenheit in Celcius ?
104 - 32 = 72
72 / 9 = 8
8 * 5 = 40
Answer: 40 C
What is 15 Celcius in Fahrenheit ?
15 / 5 = 3
3 * 9 = 27
27 + 32 = 59
Answer: 59 F
What is minus 10 Celcius in Fahrenheit ? (Notice that is below zero in Celcius, so expect an answer below 32 F)
-10/5 = -2
-2 * 9 = -18
-18 + 32 = 14
Answer: 14 F
Notice that when converting Celcius into Fahrenheit we can do: 1.8 C + 32 = F
When converting Fahrenheit into Celcius we can do: (F - 32) / 1.8 = C
Where C is the temperature in Celcius and F is the temperature in Fahrenheit. In other words using algebra for this conversion.
Notice also that (9/5) = 1.8
68 - 32 = 36
36 divide 9 = 4
4 x 5 = 20
The answer is 20 Celcius. (68 Fahrenheit)
You have missed out an important bit of the calculation. You were supposed to subtract 32 first.
This sort of conversion involves an add/subtract bit of it which makes it a more complicated conversion than the others.
Try again?
With temperature we can do Fahrenheit and Celcius (or Centigrade - which does your course use out of interest?).
To convert into Centigrade from Fahrenheit you do the following:
start with degrees F subtract 32 then divide by 9 and multiply by 5
To convert the other way it is:
start with centigrade divide by 5 multiply by 9 then add 32.
So for example: 32F is the temperature at which water freezes. Subtract 32 then divide by 9 and multiply by 5.
Since we get zero the division and multiplication has no effect. We get 0C, or just zero Celcius/Centigrade.
Another example: 50F subtract 32 and we get 18. Then divide by 9. We get 2. Then mutiply by 5. We get 10.
Answer: 10C or 10 Celcius or 10 centigrade.
What is 68 Fahrenheit in Celcius ? (From now on I will call it Celcius to save me time typing both versions.)
Updated explanation concerning: Cubic metres, litres, decilitres, centilitres and millilitres (or cubic centimetres).
A cubic metre is defined as being the volume of 100cm x 100cm x 100cm.
A litre can be regarded as the volume of a 10cm x 10cm x 10cm cube of something.
A decilitre is one tenth of a litre that is to say 10 times smaller than a litre and is 100 millilitres.
A centilitre is one hundredth of a litre that is to say 100 times smaller than a litre and is 10 millilitres.
A millilitre is one thousanth of a litre that is to say 1000 times smaller than a litre. One cubic centimetre = one millilitre
Although the millilitre is the same as a 1cm x 1cm x 1cm cube in volume notice that you cannot do this
with a centilitre or a decilitre without an irrational decimal to make the unit length of such a cube despite this the total
volume is a nice round number relative to the others.
Here is a summary of the terms on the left, then how many litres that is, then the perfect cube form of it with
a side length from the cube root of the number of cubic centimetres, and in square brakets the method for getting
the side length in cm of the perfect cube:
litre = 1 litre = 10cm x 10cm x 10cm [cube root of 1000]
decilitre = 0.1 litres = 4.64158883361cm x 4.64158883361cm x 4.64158883361cm (to 11 d.p.) [cube root of 100]
centilitre = 0.01 litres = 2.15443469003cm x 2.15443469003cm x 2.15443469003cm (to 11 d.p.) [cube root of 10]
millilitre = 0.001 litres = 1cm x 1cm x 1cm [cube root of 1]
(One million millilitres make up a cubic metre notice you have to do 100 x 100 x 100 to get a million or 1000 x 1000)
(one cubic metre = 100cm x 100cm x 100cm = 1,000,000 cubic centimetres or one million cubic centimetres or one million millilitres)
Mandy: It wouldn't hurt for you to read that bit above, but the maths of volume ratios in 3 dimensions goes
rather beyond the scope of the level you are studying, but if it helps you learn it then by all means learn
it in the above style and perhaps repeat the cube root calculations using a scientific or graphics calculator.
(Needless to say in practice even an engineer does not usually need to calculate something like that to 11 d.p.
if you can only measure to 3 signifcant figures then the answer is going to have about the same accuracy)
I don't remember whether that one has anything to do with maths. I think you should leave maths to this site.
That one seemed to me to be about dyslexia with a whole load of sub forums for dyslexic management issues.
For now I think you ought to bring up anything about maths that you want me to discuss.
I have just looked again at the website I suggested and I can't see any sub forum in there to do with maths.
However there is a massive load of information about dyslexia.
How well are you understanding the conversion topic?
Mandy: Do you want to carry on with conversion or go back to one of the other everyday maths topics?
I managed to get some good academic qualifications in maths without them, but I did in fact see a website after reading the above
post where yes apparently there are two types of division in discrete mathematical situations where you have whole numbers of an
item and you are describing the difference I think between two ways of thinking about a division, but where numerically speaking
it does not really matter as far as I could tell. If you divide 24 by 6 you get 4. Both of these numbers could be the number of items
in the group or the number of groups themselves. Understanding the real thing that is being thought about is important in division
perhaps in primary school teaching, but I would not wany to teach the primary school child confusing terms. I wonder whether in
Sets and Groups in university teaching it becomes relevant to a integer based or discrete maths topic.
There are some concepts in those sorts of "foundations" maths modules that do have this sort of feature.
For instance "parity" being even or odd being a useful term in some proofs in a proof where you need to refer to both together and
wish to keep the proof concise. So yes terms like that can sometimes be introduced like that to help the clarity of a maths topic.
If you have 26 letters of the alphabet then each of these has 26 possible values.
If my calculations are correct then you have 11881376 (26^5) values with a 5 letter word converted.
Unfortunately if you want to squash these into only 6 digits from 0 to 999999 using only numbers then there are only 1000000
numbers available to use. This needs to be just under 12 times higher to work.
If the length of the data is unbounded then obviously it is not possible because you need unlimited digital space to store it.
I think you would have to set a limit of 4 letters for a word and then map that to a set of 6 numeric digits.
This would then at least be possible, but programming that could be confusing.
However in the case I have stated of using a maximum of 4 alphabetic letters for 6 numeric digits it can be done.
Yes I think my answer was essentially a "base 100" system that I had in mind. I don't really consider it possible otherwise.
However if there were for instance a check digit after each proper digit in the original 12 digit code then it would be possible,
but we need exact details of what and how the 12 digit number was created to be able to do this.
If it does have to be secure then it would not be possible for zxcvbnm123 to post the details so we could have a problem here !
I will leave it to others from here, so feel free to take over bobbym, I think this might be more your area.
How many possible values of the 12 digit number could there be ?
If it is an integer from 0 to 999999999999 then there are 10^12 that is one billion possible numbers.
If you replace them with an integer from 0 to 999999 there are only 10^6 or only one million possible numbers.
It would be impossible to do this without information loss.
However I suppose if your encryption means that the 12 digit number is not fully used then it is possible,
provided there is deliberate redundancy in the digit usage.
Without more details of the original number it would not be possible to program this because otherwise you
would map two 12 digit numbers onto the same 6 digit number and not know how to restore the original.
There is a topic I studied about 11 years ago called "graphs networks and design" in which the design part
included a topic on codes. In this topic you could pad a set of digits out with extra information do that you could
both check and in some cases correct digits and still have them coded in some way so the information was not
obvious to someone gaining the information who should not do so. I do not have the book handy, but it might
be about somewhere. I think maybe the 12 digit number has deliberate padding of this kind, but one would need
to know how that had been done in order to decode it back to the 6 digit form. Otherwise it would not be a good code.
You can of course try to discover the code, but I presume you wrote it?
You have posted a number which contains 12 digits which I presume are from the set {0,1,2,3,4,5,6,7,8,9}
If you are trying to put that into 6 digits then you would have to do something like put them into pairs somehow such as
With 123456789123 put that into 12,34,56,78,91,23
Then find a way of mapping all of the numbers from 00 to 99 (100 of these in total) onto characters to replace them.
You would then need 100 different individual characters to do this !!
There are enough ASCII characters to do this. I do not have up to date knowledge of this because my hobbyist knowledge of the
ASCII character character set comes from about 2 to 3 decades ago. (It was first used in 1963 according to wikipedia)
With the letters of the alphabet you have 26 characters which can be lower or upper case. (52 in total)
You do of course have the 10 digits. That makes 62. Then there are punctuation characters - I'm not sure how many exactly.
With the ASCII set there are 128 characters. It might be an idea to google the term to find out about it.
So perhaps some mapping from the 100 double digit numbers to these could be worked out.
You could then map from the 12 digits to your chosen character set and then back again without losing information.
Worked example based upon the question about calculating someone's height from feet and inches into metric form.
A person is said to be 5 foot 9 inches tall exactly.
(1) How many inches is the person is just using inches ? (Hint: 12 inches equals one foot.)
5 x 12 = 60
Then 60 + 9 = 69
The person is 69 inches tall.
(2) Now convert this into centimetres. (Remember: 1 inch = 2.54 cm)
69 x 2.54 = 175.26 cm
(3) Now convert this into metres.
175.26 / 100 = 1.7526 m
(4) You could also put this in the form of metres and centimetres.
1 metre and 75 cm to the nearest cm.
(3) Convert 15cm into inches.
(4) Convert 1 foot into metres
Obviously the question (3) can be worked out more or less with a 15cm ruler.
An exact answer would be 15 / 2.54 = 5.91 inches to 2 d.p.
For question (4) we need to convert the foot into inches so 12.
Then take 12 and convert into cm. 12 x 2.54 = 30.48 cm
Then this needs to be in metres so 30.48 / 100 = 0.3048 m
I have decided to leave full accuracy in my question (4) answer.
You had 4 - 4.5 in effect.
This is -0.5
when you square that you get 0.25 then take the square root and you get 0.5 and the sign is removed.
So the assumption "if you square something then take the square root then the two cancel so you get the thing you started with"
has an exception caused by the fact that you either must not square a negative input to cause this problem - or must guarantee that the
sign is preserved always. If no such guarantee can be made the assumption is flawed.
In practice it is usually when you are trying to formally prove something that this is a problem because of the degree of rigour
expected, and because of the fact that you will want to consider a wide range of cases.
I am rather reluctant to say "you can never take the negative square root" because the quadratic formula and other results in maths
do depend upon the negative square root and sometimes real problems do need both solutions to a quadratic, but I cannot think
of any of the top of my head (some A level problems did need both solutions, or all 3 order 3 polynomial solutions etc.).