Question

In a geometric distribution, P=1/57
Calculate the following distribution:

Answer 1

\[P(13\le x<70).\]

Answer 2

You can't order me!

Answer 3

Am not ordering you :P That's my username... But I would like an answer if anyone can help :)

Answer 4

I know, I love Ema!!

Answer 5

I lv J.K Rowlin

Answer 6

Lol, Harry Potter is awesome! But can anyone help with the distribution?

Answer 7

I can, but I haven't reviewed probability distribution since soo long..

Answer 8

Ah OK, I just don't know the steps... If I have a good formula, it'd help a lot! I can't find one, however...

Answer 9

yh, it's different from binomial distribution, x (14, 13, 12.....69..), P( x attempts)

Answer 10

\[P*(1-p)^{x-1}\]

Answer 11

What would X be?

Answer 12

X is the number of attempts

Answer 13

Yes, but it would be between 13 and 70...

Answer 14

find the cummulative distribution between 13 and 70, subtract it from 1

Answer 15

P(X=k) = p(1-p)^(k-1), this tells the propability that X=k, and p is a parameter. If u want to know the propability for P(13<=x<70) its P(X=13)+P(X=14)+...+P(x=69)

Answer 16

Yes, I know that... but what's the short method in calculating with distribution without having to add all of them together?

Answer 17

p = the cummulative distribution function of x

Answer 18

Yup... I know that too... Um.. but there are like 57 distributions that I have to calculate... is there a formula for shortening it?

Answer 19

P* (1-(c.d.f)^x-1

Answer 20

P(X=13)+P(X=14)+...+P(x=69) = p(1-p)^12 + ...+ p(1-p)^68. = p(1-p)^12 (1+(1-p)+(1-p)^2 +...+(1-p)^56), where the sequence 1+(1-p) + ... is a geometric sequence, for which u can apply the formula of a sum of geometric sequence

Answer 21

So, how would you fill in the formula with P=1/57 with those parameters?

Answer 22

Easier way is to apply the cumulative distribution function for geometric propability distribution. Now CDF(k) = P(x<= k), so P(13<= x < 70) = CDF(69)-CDF(12). And for geometric propability distribution: CDF(k)=1-(1-p)^k.

Answer 23

What's the method to calculate the cumulative distribution function though? I know the method of using combinations and timesing them by the things you need to get it for let's say combination of (6,1) * 1/57^1 * 56/57^5.... but how do you do that for a large sum of distributions?

Answer 24

I gave you the cumulative distribution function for geometric distribution its CDF(k) = 1-(1-p)^k and it answers to the question P(x<=k).

It can be calculated by \[\sum_{n=1}^{k}p(1-p)^{n-1}\]

Answer 25

what is the general formula for a geometric dist func ?

Answer 26

P(x<=k) = CDF(k)=1-(1-p)^k, http://en.wikipedia.org/wiki/Geometric_distribution

Answer 27

Yup, I know that's the formula... but I don't know what n is

Answer 28

n is the index of the sum, but to your original question i already gave an answer: "Now CDF(k) = P(x<= k), so P(13<= x < 70) = CDF(69)-CDF(12). And for geometric propability distribution: CDF(k)=1-(1-p)^k"

Using the above formula the answer CDF(69)-CDF(12) with parameter p = 1/57 is 1-(1-1/57)^69 - (1-(1-1/57)^12 = (1-1/57)^12 - (1-1/57)^69 which is approximately 0.51, which is a reasonable answer to a propability question, as its between zero and one.

Answer 29

Thanks so much! Yes, that's the answer at the back of the book. I can see now why you got it.