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Anettt [7]
3 years ago
8

Part of the analysis we routinely do with datasets is to identify whether or not any of the variables included are Binomial or P

oisson in nature.
Discuss why it can be helpful to do this?
Mathematics
1 answer:
madreJ [45]3 years ago
5 0

Answer:

This is useful to choose which calculation to perform.

Step-by-step explanation:

1) Firstly, let's consider that the Binomial Distribution tends to the Poisson Distribution given certain conditions:

n\rightarrow \infty, p\rightarrow 0, \lambda =np

Roughly, they tend to the same value.

2) The Binomial Probability is calculated through this formula:

Binomial: P(X=x)=\binom{n}{x}p^{x}(1-p)^{n-x}

Poisson Distribution this way:

Poisson:P(X=x)=\frac{\lambda^{x} e^{-\lambda }}{x!}

3) If we plug

p=\frac{\lambda }{n}

In the Binomial formula, given an "n" a very large quantity we'll have a closer outcome to Poisson.

P(X=x)=\binom{n}{x}\left ( \frac{\lambda }{n} \right )^{x}(1-\frac{\lambda }{n})^{n-x} \approx \frac{\lambda^{x} e^{-\lambda }}{x!}

4) This is useful especially due to the convenience of calculating.

Because operating with exponentials and factorials, is hard and sometimes 'n' and 'p' may also be unknown, and sometimes the known parameter is the Mean.

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<h3>Distributive Property</h3>

The distributive property lets you group or ungroup terms using parentheses. It lets you multiply an external factor by every term in parentheses, expressing the result as a sum:

... a(b + c) = ab + ac . . . . . . factor <em>a</em> multiplies the terms <em>b</em> and <em>c</em>

and it lets you remove a common factor from different terms, putting that factor outside parentheses:

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<h3>Homework</h3>

20. To do this problem, you need to eliminate the parentheses using the distributive property. Then, "collect terms," which is another application of the distributive property. The external factor outside the parentheses is (-2/3). Multiply that by each term in parentheses, and add the results.

After that, recognize that c is a factor of two of the terms. Add their coefficients to simplify that sum to one term.

-\dfrac{2}{3}(12c-9)+14c=\left(-\dfrac{2}{3}\right)(12c)+\left(-\dfrac{2}{3}\right)(-9)+14c=-8c+6+14c\\\\=c(-8+14)+6\\\\=6c+6

22. You are to evaluate the expression with x=2 two ways: as is, and after you simplify it by combining like terms.

<u>As is:</u>

-8x+5-2x-4+5x\\=-8(2)+5-2(2)-4+5(2)\\=-16+5-4-4+10\\=-9

<u>Simplified:</u>

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I prefer simplifying the expression first. The number of calculations and chances for error are reduced.

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<u>First Expression:</u>

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<u>Second Expression:</u>

7x^2+7y +4x^2-4y=(7+4)x^2+(7-4)y=11x^2+3y

The simplified forms of the expressions are identical, so we conclude the expressions are equivalent.

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