Math 132A

Normal Approximation for Binomial Distributions

Binomial distribution example

Suppose we flip eight fair coins. What is the probability that there will be exactly three heads?

What is the probability that there will be at most three heads?

Probability of at most three heads

Another example

Suppose we flip twenty fair coins. What is the probability that there will be at most eight heads?

Probability of at most eight heads

Using density histogram

Normal approximation

• We have \(X \sim \operatorname{Binom}\left(20, \frac{1}{2}\right)\)
  • \(\operatorname{E} X = 10\)
  • \(\sigma_X = \sqrt{5}\)
• Approximate by \(Y \sim \operatorname{N}\left(10, \sqrt{5}\right)\)
• \(\operatorname{P}(X \le {\color{red}8}) \approx \operatorname{P}(Y \le {\color{red}8.5})\)
• \(\operatorname{P}(X \le {\color{red}8}) \approx 0.2517\)
• \(\operatorname{P}(Y \le {\color{red}8.5}) \approx 0.2512\)

\(\operatorname{P}(X \ge 6)\)

Normal approximation for \(\operatorname{P}(X \ge 6)\)

• \(X \sim \operatorname{Binom}\left(20, \frac{1}{2}\right)\)
• \(Y \sim \operatorname{N}\left(10, \sqrt{5}\right)\)
• \(\operatorname{P}(X \ge {\color{red}6}) \approx \operatorname{P}(Y \ge {\color{red}5.5})\)
• \(\operatorname{P}(X \ge {\color{red}6}) \approx 0.9793\)
• \(\operatorname{P}(Y \ge {\color{red}5.5}) = 1 - \operatorname{P}(Y \le {\color{red}{5.5}}) \approx 0.9779\)

\(\operatorname{P}(7 \le X \le 14)\)

Approximation for \(\operatorname{P}(7 \le X \le 14)\)

• \(\operatorname{P}({\color{red}7} \le X \le {\color{red}14}) \approx \operatorname{P}({\color{red}6.5} \le Y \le {\color{red}14.5})\)
• \(\operatorname{P}({\color{red}7} \le X \le {\color{red}14}) \approx 0.9216\)
• \(\operatorname{P}({\color{red}6.5} \le Y \le {\color{red}14.5}) = \operatorname{P}(Y \le {\color{red}14.5}) - \operatorname{P}(Y \le {\color{red}{6.5}}) \approx 0.9192\)

Continuity Correction

• \(\operatorname{P}(X \le {\color{red}8}) \approx \operatorname{P}(Y \le {\color{red}8.5})\)
• \(\operatorname{P}(X \ge {\color{red}6}) \approx \operatorname{P}(Y \ge {\color{red}5.5})\)
• \(\operatorname{P}({\color{red}7} \le X \le {\color{red}14}) \approx \operatorname{P}({\color{red}6.5} \le Y \le {\color{red}14.5})\)

• \(\operatorname{P}(X = {\color{red}9}) \approx \operatorname{P}({\color{red}8.5} \le Y \le {\color{red}9.5})\)

• Extend the region by 0.5 on each end.

Another example

In certain country, 14% of the population has a college degree. If we randomly select 120 members of the population, what is the probability that at least 20 of them will have a college degree?

\(X \sim \operatorname{Binom}(120, .14)\)

\(Y \sim \operatorname{N}\left(120\cdot 0.14, \sqrt{120\cdot 0.14\cdot 0.86}\right) = \operatorname{N}\left(16.8, 3.8010525\right)\)

\(\operatorname{P}(X \ge 20) \approx \operatorname{P}(Y \ge 19.5) = 1 - \operatorname{P}(Y \le 19.5)\)

\(z = \frac{19.5 - 16.8}{3.8010525} = 0.7103296\)

\(\operatorname{P}(X \ge 20) \approx 1 - 0.7611 = 0.2389\)

Does not always work

The rule

Binomial variable
\(X \sim \operatorname{Binom}(n, p)\)
can be approximated by
\(Y \sim \operatorname{N}\left(np, \sqrt{np(1-p)}\right)\)
if both \(np\) and \(n(1-p)\) are large enough.