Examples

Binomial estimation: 10,000 flips

Flip a fair coin 10,000 times. Write $H$ for the number of heads.

Estimate the probability that $4850<H<5100$.

Solution

(1) Check the rule of thumb: $p=q=0.5$ and $n=\mathrm{10,000}$, so $npq=2500\gg 10$ and the approximation is effective.

---

(2) Now, calculate needed quantities:

$$\begin{array}{c}\mu =E[X_i]\gg \gg \mu =0.5\gg \gg n\mu =5000\\ \\ {\sigma }^2=\mathrm{Var}[X_i]\gg \gg \sigma =0.5\gg \gg \sigma \sqrt{n}=50\end{array}$$

---

(3) Set up CDF:

$$F_H(h)=\mathrm{\Phi }\left(\frac{h-5000}{50}\right)$$

---

(4) Compute desired probability:

$$\begin{array}{c}P[4850<H<5100]=F_H(5100)-F_H(4850)\\ \\ \gg \gg \mathrm{\Phi }\left(\frac{100}{50}\right)-\mathrm{\Phi }\left(\frac{-150}{50}\right)\gg \gg \mathrm{\Phi }(2)-\mathrm{\Phi }(-3)\\ \\ \gg \gg \approx 0.9772-(1-0.9987)\gg \gg 0.9759\end{array}$$

Summing 1000 dice

Suppose 1,000 dice are rolled.

Estimate the probability that the total sum of rolled numbers is more than 3,600.

Solution

(1) Let $X_i$ be the number rolled on the $i^{th}$ die.

Let $S={\sum }_{i=1}^n{X}_i$, so $S$ sums up the rolled numbers.

We seek $P[S\ge 3600]$.

---

(2) Now, calculate needed quantities:

$$\begin{array}{c}\mu =E[X_i]\gg \gg \mu =7/2\gg \gg n\mu =3500\\ \\ {\sigma }^2=\mathrm{Var}[X_i]\gg \gg \sigma =\sqrt{\frac{35}{12}}\gg \gg \sigma \sqrt{n}=\sqrt{\frac{35000}{12}}\end{array}$$

---

(3) Set up CDF:

$$F_S(s)\approx \mathrm{\Phi }\left(\frac{s-3500}{\sqrt{\frac{35000}{12}}}\right)$$

---

(4) Compute desired probability:

$$\begin{array}{c}P[S\ge 3600]=1-F_S(3600)\\ \\ \gg \gg \approx 1-\mathrm{\Phi }\left(\frac{100}{54.01}\right)\gg \gg 1-\mathrm{\Phi }(1.852)\approx 0.968\end{array}$$

Continuity correction of absurd normal approximation

Let $S_n$ denote the number of sixes rolled after $n$ rolls of a fair die.

Estimate $P[S_{720}=113]$.

Solution

We have $S_n\sim \mathrm{Bin}(\mathrm{720,1}/6)$, and $np=120$ and $\sqrt{npq}=10$.

The usual approximation, since $Z$ is continuous, gives an estimate of 0, which is useless.

Now using the continuity correction:

$$\begin{array}{c}P[113\le S_{720}\le 113]\\ \\ \approx \mathrm{\Phi }\left(\frac{113+0.5-120}{10}\right)-\mathrm{\Phi }\left(\frac{113-0.5-120}{10}\right)\\ \\ \gg \gg \mathrm{\Phi }(-0.65)-\mathrm{\Phi }(-0.75)\gg \gg \approx 0.0312\end{array}$$

The exact solution is 0.0318, so this estimate is quite good: the error is 1.9%.