What exactly is a normal approximation to the binomial?

So let me get this straight. Suppose 63% of people say they don't like lemons, in a sample of 1200.

How does the idea of sample proportions come into play here?

"If N is large enough, the distribution of X ~ B(n,p) can be well approximated by the normal distribution with the same mean and variance, and so too can the sample proportion"

I don't understand exactly what a sample proportion is.

If X is a binomial distribution, then it has a unique way of calculating the mean, variance, and standard deviation. Then we also can calculate the mean, variance and standard deviation of the sample proportion. Why do we need to calculate the same thing so many times? Here is a statistic "58% of baseball players like the colour red, in a sample of 1200". This is a binomial distribution (They can either like it or they cannot). What do sample proportions have to do with it? Why is there so many ways of calculating the variance / standard deviation / mean?


1 Answer 1


Hints and Comments:

You seem to be mixing up two different limiting properties.

(1) As the sample size $n$ of a survey becomes large the sample proportion of those who like lemons tends to become ever closer to the population probability that a randomly chosen person likes lemons [That is based on the Law of Large Numbers.]

(2) The sum or average of a large number $n$ of random variables (with finite variances) tends to be nearly normal. [That is based on the Central Limit Theorem.] A binomial distribution $\mathsf{Binom}(n, p)$ is the sum of $n$ Bernoulli (0-1) random variables, and so the CLT applies to allow reasonable normal approximations of binomial distributions, especially when $n$ is large and $p$ is not too near 0 or 1 (near 1/2 is best).

Maybe that is enough to help your sort out your Question. If not, please deconvolve the LLN and CLT issues an ask about them separately.

Sample proportion: You interview $n=1000$ people and find that $X=331$ of them like lemons. Then the sample proportion $\hat p = X/n$ is the sample proportion. For large $n$ $\hat p$ is close to the population proportion $p$ who like lemons.

Finding the mean: Suppose $X \sim \mathsf{Binom}(n=3,p=.5).$ Then the by definition $$\mu_X = E(X) = \sum_{k=0}^3 kP(S=k) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 12/8 = 3/2.$$ For the binomial distribution only it turns out that there is a convenient formula: $E(X) = np = 3(1/2) = 3/2.$ Somewhat similarly, one can show that $Var(X) = np(1-p).$

Some other families of distributions have analogous special formulas for expectations and variances. You can take the attitude that there are confusingly many of them, or you can be glad that when special formulas exist for the mean and variance of a type of distribution they save a lot of work.

Suppose 63% of 1200 randomly sampled people (that's 756 of them) like lemons. Then what is the question? The best point estimate of the population proportion $p$ who like lemons is $\hat p = 756/1200 = .63.$ Of course, $\hat p$ isn't exactly $p$. You may wonder how far off the estimate might be.

Using the normal approximation to the binomial distribution, one can find a 95% confidence interval (CI) for $p.$ The simplest form of CI (OK for $n$ over 1000.) is $\hat p \pm 1.96\sqrt{ \frac{\hat p(1-\hat p)}{1200} }.$ That computes to the interval $(0.658, 0.702),$ which has $\hat p$ as its center.

  • $\begingroup$ I appended a confidence interval to my answer. No specific objective mentioned in your question. If you need something else, please specify. Maybe one of us can help. $\endgroup$
    – BruceET
    Sep 10, 2017 at 16:44
  • $\begingroup$ Thanks for that, it was very helpful $\endgroup$ Sep 21, 2017 at 4:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .