Using central limit theorem to estimate probability

Question

I roll a biased dice 9000 times. the probability of seeing 1,2,3,4,5,6 are $\frac{1}{3}, \frac{1}{12}, \frac{1}{12}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}$ respectively. I need to use the Central limit theorem to estimate the probability that the total number of 1s that I see is within [2970,3040].

So far, I only know the fact that the random variables Xi of of CLT are each rolls. I'm not sure how I'm supposed to apply the information given. Please help this hopeless soul :((

Answer 1

Define $X_i$ as the number of 1's on the i's toss. $X_i=1$ if the $i^{th}$ toss is 1, and $X_i=0$ if the $i^{th}$ toss is not 1.

By central limit theorem, we can assume $X_i's$ have normal distribution with $E(X)=9000\times\frac{1}{3}=3000$.

$Var(X)=9000\times(\frac{1}{3}-(\frac{1}{3})^2)=2000$

$\sigma=\sqrt{Var(X)}=\sqrt{2000}$

Let $T=\frac{\bar{X}-\mu}{\sigma}$, and $T$~$N(0,1)$.

So we want to calculate:

$P(2970<X<3040)=P(\frac{2970-3000}{\sqrt{2000}}<T<\frac{3040-3000}{\sqrt{2000}})$

Using R studio, with command:

pnorm((3040-3000)/sqrt(2000))-pnorm((2970-3000)/sqrt(2000))

We get $P(2970<X<3040)=P(\frac{2970-3000}{\sqrt{2000}}<T<\frac{3040-3000}{\sqrt{2000}})=0.5632858$

Answer 2

For $i=1$ to $n$, where $n=9000$, define random variable $X_i$ by $X_i=1$ if we roll a $1$ on the $i$-th roll, and by $X_i=0$ otherwise.

Let $Y=X_1+X_2+\cdots+X_n$. Then $Y$ is the sum of a large number of independent identically distributed "nice" random variables. So $Y$ has a reasonably close to normal distribution. (We have used an informal version of the Central Limit Theorem. The actual distribution of $Y$ is binomial, $n=9000$, $p=1/3$.)

I will assume that you know that the mean of $Y$ is $(1/3)(9000)$, and thst the variance of $Y$ is $(1/3)(2/3)(9000)$. So $Y$ has standard deviation $\sqrt{2000}$.

Finally, we need to calculate the probability that $$2970\le W\le 3040,$$ where $W$ is normal with mean $3000$ and standard deviation $\sqrt{2000}$. Software will do the job. But to do it the old-fashioned way, we find $$\Pr\left(\frac{2970-3000}{\sqrt{2000}}\le Z\le \frac{3040-3000}{\sqrt{2000}}\right),$$ where $Z$ is standard normal. Then the calculation can be completed using tables of the standard normal.

Answer 3

The probability of seeing a one in a single roll is $\frac{1}{3}$. The probability of $\texttt{not}$ seeing a one in a single roll is $1-\frac{1}{3}=\frac{2}{3}$.

Each roll i is bernoulli distributed. Let´s denote this random variable as $X_i$.

Therefore the sum is binomial distributed as $\sum X_i=Y\sim Bin(n,\frac{1}{3})$.

Now the approxmation is in general:

$P(Y\leq y)=\Phi\left(\frac{y-\mu}{\sigma}\right)$, where $\Phi(\cdot )$ is the cdf of the standard normal distribution.

$P(2970\leq Y\leq 3040)=P(Y\leq 3040)-P(Y\leq 2969)$

And $\mu=\frac{1}{3}\cdot 9000=3000$, $\sigma=\sqrt{n\cdot p \cdot (1-p)}=\sqrt{9000\cdot\frac{1}{3}\cdot (1-\frac{1}{3})}=\sqrt{2000}$

$P(Y\leq 3040)=\Phi\left(\frac{3040-3000}{\sqrt{2000}}\right)=\Phi(0.8944)$

$P(Y\leq 2969)=\Phi\left(\frac{2969-3000}{\sqrt{2000}}\right)=\Phi(-0.6931$)

You can use this table to evaluate the corresponding values (probabilities).

Then substract the second one from the first one.