MHB Binomial distribution regarding: (≤, >, etc.)

AI Thread Summary
The discussion revolves around calculating probabilities for a binomial random variable with parameters p = 0.2 and n = 20. The user initially struggles with understanding cumulative probabilities from the binomial table, particularly for P(X ≤ 3) and P(X = 6). Clarifications indicate that cumulative values must be used to derive specific probabilities, such as P(X = 6) being calculated as the difference between cumulative probabilities. For P(X > 10), the correct approach is to subtract the cumulative probability up to 10 from 1. The final point emphasizes the importance of correctly defining the range when calculating probabilities, leading to the correct result for P(6 ≤ X ≤ 11).
iamlorde
Messages
6
Reaction score
0
Question is as follows:

Let X be a binomial random variable with p = 0 2 .
and n = 20. Use the binomial table in Appendix A to determine
the following probabilities.
(a) P(X ≤ 3) (b) P(X > 10)
(c) P(X = 6) (d) P(6 ≤ X ≤ 11)

NK7cCqq.gif


(a) = 0.4114 is the answer. Yet all I see from this answer is that X is simple equal to "0.4114". If it is "X ≤ 3" shouldn't "0.2061", "0.0692", and "0.0115" contribute to the answer somehow because they are "<" smaller than 3?

I feel like I may be missing a fundamental element here. How do I proceed with these in general? My logic seems flawed on this matter.

For example: (c) = 0.9133-0.8042=0.1091; how is this possible. Why isn't this straight out "0.9133", the value directly next to #6?

Please enlighten me on this matter. Thank you.
 
Mathematics news on Phys.org
The numbers in the column of your table are cumulative, and so for a) you would simply read from the table to get:

a) $$P(x\le3)=0.4114$$

And for c), you would do the following:

c) $$P(x=6)=P(x\le6)-P(x\le5)=0.9133-0.8042=0.1091$$

How do you now suppose you would do parts b) and d)?
 
So, I would think,

(b) P(X > 10) = [!not](everything up to and including 10) = 1-0.9994 = 0.0006

(d) P(6 ≤ X ≤ 11) :

X is between 6 and 11, AND it includes them both, so I choose one above 11, so 12 which is: 1.0000. Then I choose one below 6, so 5, so that I can include 6. Hence: 0.8042.

Now I subtract: 1 - 0.8042 = 0.1958 (YET THIS IS FALSE)
My solution says: 0.1957?
 
Your solution for (d) is almost correct. The problem is that you've calculated $\mathbb{P}(6 \leq X \leq 12)$ because you've also included $12$. We have

$$\mathbb{P}(6 \leq X \leq 11) = \mathbb{P}(X \in \{6,7,8,9,10,11\}) = \mathbb{P}(X \leq 11) - \mathbb{P}(X \leq 5) = 0.1957$$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top