Hard probability question

**"A dice it thrown until the sum of the outcomes is greater than 300. Find, approximately, the probability that at least 80 experiments are required."**

I have the suspicion that if

X = number of throws required until sum>300

then X has normal distribution with mean 85 and ... some variance http://www.thestudentroom.co.uk/imag...olondollar.gif

How could I prove that?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ... :(

Re: Hard probability question

I doubt it's normal. We know that between 50 and 300 experiments are required, and the expected number of experiments is about 300/3.5, 85.7.

Edit: Between 51 and 301 (if you are considering "strictly greater than 300").

Re: Hard probability question

Quote:

Originally Posted by

**smalltalk** **"A dice it thrown until the sum of the outcomes is greater than 300. Find, approximately, the probability that at least 80 experiments are required."**
I have the suspicion that if

X = number of throws required until sum>300

then X has normal distribution with mean 85 and ... some variance

http://www.thestudentroom.co.uk/imag...olondollar.gif
How could I prove that?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ... :(

Hi Smalltalk,

Let's say the outcome of one die throw is $\displaystyle X_i$. Then you want to find $\displaystyle \Pr(\sum_{i=1}^{80} X_i \le 300)$. For brevity, let's say $\displaystyle Y = \sum_{i=1}^{80} X_i$.

Y is approximately normally distributed, by the Central Limit Theorem. You need to compute the mean of Y (it's not 85), and you need to find its standard deviation before you can look up the sought-for probability in a table.

Hint: $\displaystyle var(Y) = \sum_{i=1}^{80} var(X_i)$.

Can you find the variance of $\displaystyle X_i$?