Estimate the probability that, for this to happen,
- more than 210 tosses are required
- less than 190 tosses are required
- 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.
Let $X[1]...X[N]$ 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 $P(S_{180}<700<S_{210})$