Normal approximation to the binomial dis- tribution. Suppose the distribution for the ran- dom sample is a Bernoulli distribution with pa- rameter p, and, as usual, let q = 1 − p. In other words we have n independent Bernoulli trials The sample mean and X1,X2,...,Xn, each Xi having a probability p of the central limit theorem success, that is, P (Xi=1) = p, and probability q of Math 218, Mathematical Statistics failure, that is, P (Xi=0) = q. D Joyce, Spring 2016 Recall that the mean of the Bernoulli distribution is µ = p and its variance is σ2 = pq. The distribution of the sample mean. You saw When each of the independent random variables last semester that the sample mean is approxi- X1,X2,...,Xn is a Bernoulli distribution, then the mately normally distributed according to the cen- Pn sum Sn = i=1 Xi is a binomial distribution with tral limit theorem. First we’ll review that, then see parameters n and p. This binomial distribution has how it can be used to approximate the binomial 2 mean µSn = np and variance σSn = np(1−p) = npq. distribution. By the central limit theorem, the normalized standardized sum is approximately normal. That Review of the Central Limit Theorem. Con- means that sider a random sample ∗ Sn − np S = √ n npq X = (X1,X1,...,Xn), that is, an array of n independent random vari- is approximately a standard normal distribution. We can use that fact to quickly find probabil- ables with the same distribution (each Xi is called a trial), and suppose that this distribution has mean ities related to this distribution using tables for µ = µ and variance σ2 = σ2 . The sample sum is standard normal distributions. For instance, sup- X X pose we want to know the probability of getting Sn = X1 + X2 + ··· + Xn, no more than 8 heads when flipping a fair coin 2 2 n = 20 times. That means we want to know the and it has mean µSn = nµ and variance σSn = nσ . The sample mean is probability P (Sn ≤ 8). But Sn is approximately 1 1 normal with mean µSn = np = 10 and variance X = Sn = (X1 + X2 + ··· + Xn), σ2 = npq = 5. Now, 8 is 2 less than the mean n n Sn √ and it has mean µ = µ and variance σ2 = σ2/n. of 10, and 2 is 2/ 5 = 0.8944 standard deviations, X X ∗ therefore The normalized sample mean X is the same as ∗ 8 − 10 the normalized sample sum Sn P (Sn ≤ 8) ≈ P (Z ≤ √ ) 5 ∗ X − µ Sn − nµ X = √ = √ = S∗. = Φ(−0.8944) = 0.1867. σ/ n σ n n The central limit theorem says that this normal- (That last figure comes from the standard normal ized sample mean approaches a standard normal curve areas Φ(z) on table A.3, page 673.) distribution as n → ∞. That means that the un- But we can do better if we use the so-called con- normalized sample mean X is approximately nor- tinuity correction and use 8.5 instead of 8, to get 2 2 mal with mean µX = µ and variance σ = σ /n, X 8.5 − 10 and that the unnormalized sample sum Sn is ap- P (Sn ≤ 8.5) ≈ P (Z ≤ √ ) proximately normal with mean µSn = nµ and vari- 5 2 2 ance σSn = nσ . = Φ(−0.6708) = 0.2514.

1 That last figure is very close the exact value of 0.2517.

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2