How to Find the Normal Approximation to the Binomial with a Large Sample n
If you are working from a large statistical sample, then solving problems using the binomial distribution might seem daunting. However, there’s actually a very easy way to approximate the binomial distribution, as shown in this article.
Here’s an example: suppose you flip a fair coin 100 times and you let X equal the number of heads. What’s the probability that X is greater than 60?
In a situation like this where n is large, the calculations can get unwieldy and the binomial table runs out of numbers. So if there’s no technology available (like when taking an exam), what can you do to find a binomial probability? Turns out, if n is large enough, you can use the normal distribution to find a very close approximate answer with a lot less work.
But what do we mean by n being “large enough”? To determine whether n is large enough to use what statisticians call the normal approximation to the binomial, both of the following conditions must hold:
To find the normal approximation to the binomial distribution when n is large, use the following steps:

Verify whether n is large enough to use the normal approximation by checking the two appropriate conditions.
For the above coinflipping question, the conditions are met because n ∗ p = 100 ∗ 0.50 = 50, and n ∗ (1 – p) = 100 ∗ (1 – 0.50) = 50, both of which are at least 10. So go ahead with the normal approximation.

Translate the problem into a probability statement about X.
In this example, you need to find p(X > 60).

Standardize the xvalue to a zvalue, using the zformula:
For the mean of the normal distribution, use
(the mean of the binomial), and for the standard deviation
(the standard deviation of the binomial).
So, in the coinflipping example, you have
Then put these values into the zformula to get
To solve the problem, you need to find p(Z > 2).
On an exam, you won’t see
in the problem when you have a binomial distribution. However, you know the formulas that allow you to calculate both of them using n and p (both of which will be given in the problem). Just remember you have to do that extra step to calculate the
needed for the zformula. You can now proceed as you usually would for any normal distribution.

Look up the zscore on the Ztable and find its corresponding probability.

a. Find the row of the table corresponding to the leading digit (one digit) and first digit after the decimal point (the tenths digit).

b. Find the column corresponding to the second digit after the decimal point (the hundredths digit).

c. Intersect the row and column from Steps (a) and (b).
Continuing the example, from the zvalue of 2.0, you get a corresponding probability of 0.9772 from the Ztable.


Select one of the following.

a. If you need a “lessthan” probability — that is, p(X < a) — you’re done.

b. If you want a “greaterthan” probability — that is, p(X > b) — take one minus the result from Step 4.
Remember, this example is looking for a greaterthan probability (“What’s the probability that X — the number of flips — is greater than 60?”). Plugging in the result from Step 4, you find p(Z > 2.00) = 1 – 0.9772 = 0.0228. So the probability of getting more than 60 heads in 100 flips of a coin is only about 2.28 percent. (In other words, don’t bet on it.)

c. If you need a “betweentwovalues” probability — that is, p(a < X < b) — do Steps 1–4 for b (the larger of the two values) and again for a (the smaller of the two values), and subtract the results.

When using the normal approximation to find a binomial probability, your answer is an approximation (not exact) — be sure to state that. Also show that you checked both necessary conditions for using the normal approximation.