Lesson 12: Binomial Distribution

Suppose the chance of getting a head by flipping a bent coin is $0.70$. What is the probability of getting one head in four trials?

In general, if the probability of a success for a single trial is $p$, then the chance of observing exactly $k$ successes in $n$ independent trials is given by

$$P(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{(n–k)}=\frac{n!}{k!(n–k)!}{p}^k(1-p{)}^{(n–k)}$$.

Four Conditions to check for binomial:

1. The trials are independent.
2. The number of trials, $n$, is fixed.
3. Each trial outcome can be classified as a success or failure.
4. The probability of a success $p$, is the same for each trial.

Normal Approximation of the Binomial Distribution

Let $N(\mu ,\sigma )$ be the standard notation to indicate a normal distribution with mean $\mu$ and standard deviation $\sigma$.

Rule of thumb: the binomial distribution with probability of success $p$ may be approximated by the normal distribution if the sample size $n$ is sufficiently large so that $np$ and $n(1-p)$ are both at least 10.

More specifically, using the parameters

• mean: $\mu =np$
• standard deviation: $\sigma =\sqrt{np(1-p)}$

then the following approximations hold:

• $P(X\le k)\approx P(N(\mu ,\sigma )\le k+0.5);$
• $P(X\ge k)\approx P(N(\mu ,\sigma )\ge k-0.5).$

The 0.5 is known as a continuity correction .

Example: if $n=100$ and $p=0.2$, then $np=20$ and $n(1-p)=80$ are both greater than 10, $\mu =np=20$, $\sigma =\sqrt{np(1-p)}=\sqrt{20(0.8)=4}$ and
$$0.1285=P(X\le 15)\approx P(N(20,4)\le 15.5)=0.1303.$$

As you can see, the approximation may be very rough, even if the “rule of thumb” conditions are met.

Galton Board with different sized balls

• Thyago: Doesn’t the fact that some balls are hitting each other interfere with the independence assumption? I mean, the probability of going right or left is biased by the interactions between the balls.
• Chris Beck: Lack of independence seems to be causing the drift from the normal distribution. The large balls are processed relatively independently as only 1 can fit through each condition at a time.
• David Bulger: Watching the small balls fall, you can see that they tend to gather momentum and run in diagonal paths to the left or right—that is, they don’t change direction much. This illustrates the importance of the independence assumption in the central limit theorem: if the individual random variables are not independent, then their sum may not tend to a normal distribution.

Why do YOU think the results are so different depending on the size of the balls?