A die is thrown until the first time the total sum of the face values of the die is 700 or greater. What is the probability for $n$ tosses?

Estimate the probability that, for this to happen,

1. more than 210 tosses are required
2. less than 190 tosses are required
3. between 180 and 210 tosses, inclusive, are required

We're supposed to be using the central limit theorem so I set up the inequality just for $$n$$ then manipulated to get a standardized random variable. I keep getting really weird negative numbers that don't work in the table. Not sure how to go about this one.

So far,

$$P(210 < n) =\\ P(735 < nE(x)) =\\ P(35 < nE(x)-S_n) =\\ P(-35 > S_n-nE(x)) =\\ P(-35/\sqrt{\frac{7350}{12}} > S_n-nE(x)/\sqrt{n\operatorname{Var}(x)})$$

This is where I'm stuck. Sorry if this looks confusing I don't know how to make it look better. I know this is wrong, I've been working on it for a while and I have no idea how to approach it that's why I'm asking.

• Bit hard to offer advice when you do not show your working. – David Feb 8 at 6:45
• Show exactly what you've done so far. – Henno Brandsma Feb 8 at 6:45
• This is a good example of why you should use MathJax. The question was completely unreadable, I'm not really sure my attempt to clean it up has actually helped. You should also tell us why you get each of the equalities. – Henrik Feb 8 at 7:25
• Sorry I've never heard of MathJax, this is my first time posting on here. – Katie Feb 8 at 17:57

Let $$X...X[N]$$ be the outcomes of each of the rolls. Each outcome is i.i.d. (independent and identically distributed). This is a uniform random variable where $$E[X[i]] = \frac 72$$ and $$Var[X[i]] = \frac{35}{12}$$

Suppose you roll N times, then the sum of the outcomes will be $$S_N=\sum_{i=1}^N X[i]$$

$$E[S_N] = E[\sum_{i=1}^N X[i]]=\sum_{i=1}^N E[X[i]]=\sum_{i=1}^N \frac72=\frac 72N$$

$$Var[S_N] = Var[\sum_{i=1}^N X[i]]=\sum_{i=1}^N Var[X[i]]=\sum_{i=1}^N \frac{35}{12}=\frac{35}{12}N$$

First question is asking $$P(S_{210}<700)$$. $$E[S_{210}] = 735$$ and $$Var[S_{210}] = 612.5$$

$$P(S_{210}<700) = P(\frac {S_{210}-E[S_{210}]}{\sqrt{Var(S_{210})}}<\frac {700-E[S_{210}]}{\sqrt{Var(S_{210})}})=P(z<\frac {700-735}{\sqrt{612.5}})\approx P(z<-1.414)\approx 0.0787$$

Do similar steps for 2 and 3.

2 is saying $$P(S_{190}>700)$$ and 3 is saying $$P(S_{180}<700

• Thank you. The part that was throwing me off was the varied amount of n tosses. I thought that it would mean another equality would have to be calculated. It makes sense that you just let n equal the 210 etc. tosses. – Katie Feb 8 at 18:09