Poisson's Distribution application in Radioactive decay

[AHSAN MUJTABA]

Summary:: I have been provided with the table in which N and occurence are given. I have been asked to calculate the 1. Total count 2. mean count 3. mean count.
Now, assuming our distribution described by Poisson's we need to calculate the tasks.
The Poisson's distribution is given by: $P(N,\bar N)=\frac{exp(-\bar N)\bar N^{N}}{N!}$
The expected value are calculated as, $\bar N=\Sigma P N$

As long as I understood the attached question, I have following attempt to the questions.
1. I took the sum over all N that came out to be = 89 (the total number of counts recorded)
2. I took a simple data mean with $\frac{n}{N}$ for which n=total number of data points I have right now i.e. n=12
3. I calculated the mean number of counts $\bar N$ by the fact that there are 58 trials so $\bar N=\frac{N}{no. of occurences}=\frac{89}{58}=1.5345 \frac{counts}{sec}$
4. In this part I am confussed because it asks me to calculate the expected no of occurences but by the given formulae I am unable to understand whether I should calculate $\bar N$.

 

[AHSAN MUJTABA]

And also since I don't have a count value; $N=0$ then my probability for it should be 0 but why it is not zero??

 

[Office_Shredder]

I think 0 is a possibility, but it just never occurred in those trials. Where did you get the number 89 from?

 

[AHSAN MUJTABA]

I think 0 is a possibility, but it just never occurred in those trials.Where did you get the number 89 from?

I calculated it like total N; because these are the total number of counts that are recorded but I am not sure about that.

 

[AHSAN MUJTABA]

I also think that if I want to calculate the expected value of the occurences, then I have total 58 ocurrences then I would calculated them as, $\bar O=\Sigma O P$ where P is Poisson's probability. I guess?

 

[Office_Shredder]

I don't understand what you mean when you say I calculated it like total N. Just eyeballing the table it seems like most of the time they get 6-10 particles power per second, so your estimate for the mean must be too small.

 

[AHSAN MUJTABA]

I don't understand what you mean when you say I calculated it like total N. Just eyeballing the table it seems like most of the time they get 6-10 particles power per second, so your estimate for the mean must be too small.

In the first part, they asked to calculate the total number of counts recorded, so won't that be the sum of N's mentioned in the table??

 

[AHSAN MUJTABA]

In the first part, they asked to calculate the total number of counts recorded, so won't that be the sum of N's mentioned in the table?? N is no. of counts from the table.

 

[Office_Shredder]

Can you just write down literally the sum you computed? Like 1+2+3+6+….

 

[AHSAN MUJTABA]

Can you just write down literally the sum you computed? Like 1+2+3+6+….

Yes I did the sum in just as you quoted $\Sigma N=sum(1,3,4,5,6,7,8,9,10,11,12,13)=89$ This maybe giving me the total number of counts recorded in 58 trials.

 

[Office_Shredder]

I see. To take one example, the number 7 occurred 11 different times. To calculate the mean count you need to include 7 eleven times in the average you compute.

 

[AHSAN MUJTABA]

I see. To take one example, the number 7 occurred 11 different times. To calculate the mean count you need to include 7 eleven times in the average you compute.

Ok! Are you suggesting this formula: $\Sigma \frac{N(corresponding occurence)}{total occurence}$?? I think I missing something..

 

[Office_Shredder]

To take another example, suppose you rolled a six sided die 100 times and got a 1 18 times, a 2 21 times, a 3 15 times, a 4 13 times, a 5 17 times and a 6 16 times.

What is the expected value of the die roll? You effectively just added up 1+2+3+4+5+6, which doesn't actually tell you anything about the statistics here.

 

[AHSAN MUJTABA]

Is it,
$\frac{1*18+2*21+3*15+4*13+5*17+6*16}{100}$?? According to my understanding.

 

[AHSAN MUJTABA]

In that way when I computed the total number of counts in part 1, I got 423 as an answer.

 

[Office_Shredder]

And so what do you get for the expected number in a second?

 

[AHSAN MUJTABA]

And so what do you get for the expected number in a second?

I got $7.29$ or $7.3$

 

[AHSAN MUJTABA]

This answer and logic now makes sense for me, but again the thing that bothers me is to calculate the mean count i.e. the second part of the question.

 

[Office_Shredder]

To be honest that part confuses me a little. I think it is the same number, but different units. The mean count is 7.3, the mean count rate is 7.3/second. If the sampling time was not 1 second, you would get different numeric answers.

 

[AHSAN MUJTABA]

Ok, I have now, calculated the expected number of occupations by using relation $\bar O=O\times P(N\leq 5)$
for the probability of 5 or fewer occurences, I started a series from $N=0\rightarrow 5$ and used the formula of the Poission's distribution. The answer I got is $15.312$ for the 4th part of the question.

 

[AHSAN MUJTABA]

And by the same approach, I calculated $P(N\geq 20)$ by computing $P(N\leq 20)$ then did $1-P(N\leq 20)=2.64\times10^{-5}$ which makes sense I think cause there's very little chance for $N>20$ to appear as the data says. Finally, got expected value of occurences as $0.0015$. But still I am confussed.

 

[AHSAN MUJTABA]

To be honest that part confuses me a little. I think it is the same number, but different units. The mean count is 7.3, the mean count rate is 7.3/second. If the sampling time was not 1 second, you would get different numeric answers.

I think that also makes sense to me if I take second as my unit then mean value would be $\bar N=\frac{7.3}{sec}\times sec$.