Binomial, Poisson and Normal Probability distribution help

AI Thread Summary
The discussion focuses on understanding and calculating probabilities using Binomial, Poisson, and Normal distributions in the context of specific math problems. The first question involves determining the probability of a chocolate bar containing a certain number of hazardous squares, with users seeking clarification on whether to use Cumulative or Probability Mass functions. The second question pertains to the probability of students developing a "math addiction," requiring conditional probability calculations based on course enrollment. Participants share their calculations and results, expressing uncertainty about the correctness of their methods and answers. The conversation emphasizes the importance of distinguishing between cumulative and discrete probability functions in statistical analysis.
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Binomial, Poisson and Normal Probability distribution help!

Hey everyone, I've just started a new section on probability (argh) in my math course, and unlike most other maths, I cannot cope! Anyway, I was wondering if anyone could tell me if I did the following question correctly! Any helps greatly appreciated guys! Thanks.

Q1. A chocolate factory makes gigantic “death by chocolate” chocolate bars and finds that,
in the long run, approximately 10% of the individual squares contain so much chocolate
to present a health hazard to people consuming them. Every chocolate bar contains 100
squares.
(a) Using the Binomial, Poisson and Normal distributions, write down formulas for
the probability that a single chocolate bar has at least 3 but no more than 7 deadly
squares.

(b) Use either your calculator or the Scientific Notebook functions BinomialDist, PoissonDist
and NormalDist to estimate each of these probabilities.
Does anyone know if this question would be referring to a Cumulative Probability Distribution Function, or a Probability Mass function? (Important!)

Q2.John teaches two courses in second semester, MTH1122 and MTH1030. Among
John’s students 20 percent are MTH1122 students. As a consequence of John’s
teaching technique 10 percent of the MTH1030 students and 5 percent of the MTH1122
students develop an addiction to mathematics (wishful thinking, or course, but...).
a) What is the probability that one of John’s students who has been randomly chosen
is a matheholic? (5 marks)
b) One of John’s students presents at the doctors with acute symptoms of maths addiction.
What is the probability that this student is an MTH1030 matheholic?

This question assumes:A student who takes one course, does not take the other course. There were no matheholics among the students before Burkard started teaching the course.

MY ANSWERS:

Q1. (a) Binomial: Using, n=100, p=0.10 and I calculated the formula as:

Pr(3[greater than or equal to]P[less than or equal to]7)=((100!)/(3!*97!)*((0.10)^3)*((0.90)^97) + ... + ((100!)/(7!*93!)*((0.10)^7)*((0.90)^93)

Poisson: I calculated the mean, as E(p)=np=100*0.10=10

therefore, Pr(3[greater than or equal to]P[less than or equal to]7)=([exp]^(-10)*((10^3)/3!)) + ...+([exp]^(-10)*((10^7)/7!))

Normal: Using the mean (E(P)) to get the standard deviation of P (stdev(p)=(Var(P))^(1/2)

Pr(3[greater than or equal to]P[less than or equal to]7)= [Integral: upper limit=7, lower limit=3] of ((1)/3((2pi)^(1/2))*[exp]^(-0.5(((x-10)/3)^2)

(b) For (b), I used excel, however I don't know whether this question is a 'Cumulative Probability Distribution Function' or a 'Probability Mass Function'. Anyone know?? Because, using Excel, if I use a CDF(for Binomial), I get the above probability as 0.412331, and if I use a PMF, i get the above probability as 0.204706.

Q2. For this question, it looks and sounds really easy, but I don't trust my mischievous lecturer (as this question is worth 10 marks out of a 50 mark assignment).

Firstly I set up a tree diagram:(see attached picture), where M=probability a student is a 'matheholic'

(a) For question a, I said: Pr(M)= Pr(MTH1122 student AND a matheholic) + Pr(MTH1030 student AND a matheholic)=(0.20*0.05) + (0.80*0.10)=0.09

(b) I set it up as a conditional probability question: Pr(MTH1030 Matheholic|M)= Pr(MTH1030 Matheholic AND M)/ Pr(M)=0.08/0.09=0.889

Again, ANY Help at all is greaatly appreciated, and thanks for your time!
 

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Probability mass function is the name given to the function used in discrete probability to calculate values of the form \Pr(X = a). Cumulative distribution functions are used to calculate \Pr(X \le a). CDFs exist for both discrete and continuous distributions; continuous distributions (like the normal) have density functions, which do not, by themselves, give probability.
 


So for question 2., I'd use a Cumulative Probability Distribution function ,as I am finding the probability of 'less than or equal to' and 'greater than or equal to'. Is this correct??
 


You most recent question:

His_Dudeness3 said:
So for question 2., I'd use a Cumulative Probability Distribution function ,as I am finding the probability of 'less than or equal to' and 'greater than or equal to'. Is this correct??

Yes, that's correct. Authors usually use upper-case letters to ''name'' cumulative distribution functions. If X is a random variable then we might call its c.d.f. G.
Then

<br /> \Pr(X \le x) = G(x)<br />

- the probability that X is less-than-or-equal-to some specific value x is precisely G(x). There are some intricacies involved in calculating expressions like \Pr(X &gt; x) and \Pr(X \ge x) , but those occur primarily when you deal with &lt;b&gt; discrete &lt;/b&gt; versus &lt;b&gt; continuous &lt;/b&gt; random variables and distributions. (If you&amp;#039;ve heard the names, a binomial distribution is one example of a discrete distribution, a normal distribution is one example of a continuous distribution.)
 


statdad said:
You most recent question:



Yes, that's correct. Authors usually use upper-case letters to ''name'' cumulative distribution functions. If X is a random variable then we might call its c.d.f. G.
Then

<br /> \Pr(X \le x) = G(x)<br />

- the probability that X is less-than-or-equal-to some specific value x is precisely G(x). There are some intricacies involved in calculating expressions like \Pr(X &gt; x) and \Pr(X \ge x) , but those occur primarily when you deal with &lt;b&gt; discrete &lt;/b&gt; versus &lt;b&gt; continuous &lt;/b&gt; random variables and distributions. (If you&amp;#039;ve heard the names, a binomial distribution is one example of a discrete distribution, a normal distribution is one example of a continuous distribution.)
&lt;br /&gt; Thanks for the clarification statdad!
 


So just to clarify it for me, this question we can use conditional probability right?:

Q2.John teaches two courses in second semester, MTH1122 and MTH1030. Among
John’s students 20 percent are MTH1122 students. As a consequence of John’s
teaching technique 10 percent of the MTH1030 students and 5 percent of the MTH1122
students develop an addiction to mathematics (wishful thinking, or course, but...).
a) What is the probability that one of John’s students who has been randomly chosen
is a matheholic? (5 marks)
b) One of John’s students presents at the doctors with acute symptoms of maths addiction.
What is the probability that this student is an MTH1030 matheholic?

and with this question we use Cumulative Probability Distribution function right?:

(1b) Use either your calculator or the Scientific Notebook functions BinomialDist, PoissonDist
and NormalDist to estimate each of these probabilities.
 


So are we supposed to be getting approximately the same values for the Poisson, Binomial and Normal Distributions? The answer's I got were:

BTW, before I start, I calculated the mean for the Binomial,Poission and Normal distributions using E(X)=np=100*0.1=10. And I calculated the variance using these numbers, getting Var(X)=np(1-p)=10*.9=9, and thus, the standard deviation of X=(9)^(1/2)=3.

For Binomial, Pr(3<X<7)=0.412331, I got this using Excel using the BINOMDIST function where I calculated Pr(X=3)+Pr(X=4)+...+Pr(X=7) using the Cumulative Distribution function.

For Poisson, Pr(3<X<7)=0.217457, I just did this seperately for Pr(X=3)+Pr(X=4)+...+Pr(X=7) on my graphics calculator where i used the mean as 10.

For Normal, Pr(3<X<7)=0.33022, I got this using excel, where I calculated the individual probabilities using NORMDIST= where I used X=no. of trials, mean=10 and stdev.=3. Since I am calculating [Greater than or equal to] and [Less than or equal to], I set it as TRUE (meaning I calculated using the Cumulative Distribution function). Can anyone tell me if these answers are correct?
 
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