# Normal approximation on (what it looks like) a poisson

I am self-studying inferential statistics from Larson's introductory textbook named: "Introduction to Probability Theory and Statistical Inference" (1st edition John Wiley & Sons.).

I came across an Exercise on the chapter of Central Limit Theorem, that has the following description:

A pair of dice are rolled 180 times an hour at a craps table. What is the probability that 25 or more rolls have a sum of 7 during the first hour?

My understanding is that the problem indicates that I have to approximate a Poisson random variable with the standard normal. Hence, I consider a random variable $Y$ as the sum of rolls with a 7. This way, I approximate $F_{Y}(t) = N_{Z}(\frac{t - n\lambda}{\sqrt{n\lambda}})$ and I substitute $n=1$ (since I am interested in 1 hour only), $\lambda=180$, and I solve for $t=25$. However, I get $N_{Z}(-11.55)$, which from a standard normal calculator I get a probability of $1$. The correct solution from the book is $.8643$.

Obviously, I am doing something wrong in my approach and I would like some indication on what is the wrong with my solution.

• Please add the [self-study] tag – kjetil b halvorsen May 12 '18 at 20:58

The probability of "success" in a single draw is $p=1/6$, since there are 6 possibilities out 36 to have a sum of 7. If the number of draws were constant let's say $N=180$, then this is a binomial problem, i.e. $X\sim Bin(180,1/6)$ where $X$ is the number of successes out of 180 draws. However, the problem is $N$ is stochastic and follows a Poisson distribution.

To address this, one can think about the problem differently. If $Y$ is the number of the draws in an hour in which we get a success, then it follows that $$P(Y = k) = \sum_{n=k}^\infty P(Y=k \mid N = n)P(N=n)$$ where $$Y \mid N = n \sim Bin(n,1/6)$$ and $$N \sim Poisson(180)$$

If you solve the sum above, it follows that (see link) $$Y \sim Poisson(180\times 1/6)$$ Finally, the answer is $$P(Y > 24) = 0.8428$$ which is the exact answer.

Alternatively, you approximate this using the CLT returning, where $$Y\sim N(30,30)$$

The answer for $P(Y > 24) = 0.8634$. However, since we are approximating a discrete variable using a continuous one, it is common to compute the probability of $P(Y > 24.5)$, which yields a closer answer to the exact one.

To further elaborate, let $Y_m$ denote the number of events in minute $m$, such that $Y_m \sim Poisson(0.5)$. The number of events taking place in $M$ minutes, therefore, is $Y(M)=\sum_{m=1}^M Y_m \sim Poisson(0.5M)$. Also, recall that $E \left[ Y(M) \right]=V \left[ Y(M) \right] = 0.5M$. According to the CLT, thus, we have $$Y(M)=\sum_m^M Y_m \sim N\left(0.5M,0.5M \right)$$ As a result, when $M=60$, the answer is $$P(Y>24) = 1 - \Phi \left(\frac{24 - 30}{\sqrt{30}} \right) = 0.8634.$$

• thank you for your answer. However, I would like to see how it can be approximated using the CLT. I would greatly appreciate it If you could provide an answer based on the CLT approximation. – nick.katsip May 10 '18 at 17:36
• I added a paragraph at the end. HTH – majeed simaan May 12 '18 at 20:56