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Calculate binomial coefficient casio fx82es - Office Equipment & Supplies

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6ya6ya
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Posted on Jan 02, 2017

SOURCE: how to calculate the Linear correlation coefficient

go to [2nd] [0] then select [diagnosticon] and click [ENTER] then it atomatically adds it to the LinReg(ax+b)

Posted on Jan 21, 2008

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SOURCE: Binomial distribution

Go to this page.
http://www.elderlab.yorku.ca/~aaron/Stats2022/BinomialDistribution.htm

Posted on Jan 26, 2009

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SOURCE: need help to calculate correlation coefficient

Hi,

I don't know if you still need help with this problem but in case you do follow this link:
http://www.mathstore.net/graphing-calculators/TI-calculators/how-to-calculate-the-correlation-coefficient.php

Good luck!

Posted on Jun 15, 2009

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SOURCE: binomial function in calculator fx-911ES

Hello,

The binomial probability distribution is defined as
P(r;p;n) =(nCr)(p^r)*(1-p)^(n-r)
where n is the number of trials, p the probability of success, and r the expected result.

Let n=20, r=7, p=0.15 ( I do not know wether this exemple has any meaning in the context of your problem, but you have to enter values that mean something to you. I am only showing you the key strokes

To enter 20C7 you press 20 [SHIFT][nCr]7 ;
To enter 0.15 to the power 7 you type 0.15[X to ] 7 the key is between
[x²] and [log]

To enter (1-0.15) to power 20-7, you type 0.85 [X to] 13
With [*] standing for multiplication key , and [X to] the raise to power key, the exemple above can be entered as

( 20 [SHIFT][nCr] 7) [*] ( 0.15 [X to] 7 ) [*] ( 0.85 [X to] 13 ) [=]

Here is a screen capture to show you what it looks like. However on this calculator the combination 20 [SHIFT][nCr] 7 is represented as nCr(20,7).

e74be60.jpg

Hope it helps

Posted on Nov 15, 2009

  • 7993 Answers

SOURCE: binomial function

Hello,

The binomial probability distribution is defined as
P(r;p;n) =(nCr)(p^r)*(1-p)^(n-r)
where n is the number of trials, p the probability of success, and r the expected result.

Let n=20, r=7, p=0.15 ( I do not know wether this exemple has any meaning in the context of your problem, but you have to enter values that mean something to you. I am only showing you the key strokes

To enter 20C7 you press 20 [SHIFT][nCr]7 ;
To enter 0.15 to the power 7 you type 0.15[X to ] 7 the key is between
[x²] and [log]

To enter (1-0.15) to power 20-7, you type 0.85 [X to] 13
With [*] standing for multiplication key , and [X to] the raise to power key, the exemple above can be entered as

( 20 [SHIFT][nCr] 7) [*] ( 0.15 [X to] 7 ) [*] ( 0.85 [X to] 13 ) [=]

Here is a screen capture to show you what it looks like. However on this calculator the combination 20 [SHIFT][nCr] 7 is represented as nCr(20,7).

e74be60.jpg

Hope it helps

Posted on Nov 15, 2009

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Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:


oneplusgh_15.jpg
Now to solve using the Ti 89.


Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

oneplusgh.gif

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.



oneplusgh_26.jpg


The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.



oneplusgh_18.jpg

This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.


oneplusgh_19.jpg

This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y



oneplusgh_20.jpg


This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2



oneplusgh_27.jpg



This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3



oneplusgh_28.jpg



This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4



oneplusgh_21.jpg


This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5



oneplusgh_22.jpg

This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6



oneplusgh_23.jpg

Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6












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