Normal Approximation to the Binomial for AP Statistics

By — McGraw-Hill Professional
Updated on Feb 4, 2011

Practice problems for these concepts can be found at:

Under the proper conditions, the shape of a binomial distribution is approximately normal, and binomial probabilities can be estimated using normal probabilities. Generally, this is true when np ≥ 10 and n(1 – p) ≥ 10 (some books use np ≥ 5 and n(1 – p) ≥ 5; that's OK). These conditions are not satisfied in Graph A (X has B(20, 0.1)) below, but they are satisfied in Graph B (X has B(20, 0.5))

It should be clear that Graph A is noticeably skewed to the right, and Graph B is approximately normal in shape, so it is reasonable that a normal curve would approximate Graph B better than Graph A. The approximating normal curve clearly fits the binomial histogram better in Graph B than in Graph A.

When np and n(1 – p) are sufficiently large (that is, they are both greater than or equal to 5 or 10), the binomial random variable X has approximately a normal distribution with

Another way to say this is: If X has B(n, p), then X has approximately , provided that np ≥ 10 and n(1 – p) ≥ 10 (or np ≥ 5 and n(1 – p) ≥ 5).

example: Nationally, 15% of community college students live more than 6 miles from campus. Data from a simple random sample of 400 students at one community college are analyzed.

  1. What are the mean and standard deviation for the number of students in the sample who live more than 6 miles from campus?
  2. Use a normal approximation to calculate the probability that at least 65 of the students in the sample live more than 6 miles from campus.

solution: If X is the number of students who live more than 6 miles from campus, then X has B(400, 0.15).

  1. μ = 400(0.15) = 60; σ = = 7.14.
  2. Because 400(0.15) = 60 and 400(0.85) = 340, we can use the normal approximation to the binomial with mean 60 and standard deviation 7.14. The situation is pictured below:

Using Table A, we have = 1 – 0.7580 = 0.242. By calculator, this can be found as normalcdf(65,1000,60,7.14) = 0.242. The exact binomial solution to this problem is given by 1-binomcdf(400,0.15,64) = 0.261 (you use x = 64 since P(X ≥ 65) = 1 – P(X ≤ 64)).

In reality, you will need to use a normal approximation to the binomial only in limited circumstances. In the example above, the answer can be arrived at quite easily using the exact binomial capabilities of your calculator. The only time you might want to use a normal approximation is if the size of the binomial exceeds the capacity of your calculator (for example, enter binomcdf(50000000,0.7,3250000). You'll most likely see ERR:DOMAIN, which means you have exceeded the capacity of your calculator, and you didn't have access to a computer. The real concept you need to understand the normal approximation to a binomial is that another way of looking at binomial data is in terms of the proportion of successes rather than the count of successes. We will approximate a distribution of sample proportions with a normal distribution and the concepts and conditions for it are the same.

Practice problems for these concepts can be found at:

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