Normal Approximation to the Binomial Distribution

Answering Probability Questions

With normally distributed continuous data, we can answer questions of probability by transforming raw scores to z-scores (i.e., creating a standard normal distribution) and using the unit normal table.  We can use a similar process with binomial data because binomial distributions tend to be (though are not exactly) normal in shape.  When we convert raw scores from a binomial distribution to z-scores, the result is called the Normal Approximation.  Recall that a z-distribution maintains the same shape as the original un-transformed distribution.  Because a binomial distribution is not a true normal curve, the resulting z-distribution only approximates a normal distribution. 

Two conditions must be satisfied in order to use the Normal Approximation to answer probability questions about binomial data.  As we increase n, the binomial distribution more closely approximates a normal distribution.  Before we use the normal approximation to determine probabilities, we want to be sure that the original binomial distribution is fairly normal in shape.  We can be certain of this if:   (1)  pn > 10  and  (2)  qn > 10

The normal approximation of a binomial distribution has m = pn and s = the square root of npq .

To gain maximum accuracy when using the normal approximation, we need to use the real limits of the X scores.  To find the probability of obtaining a score > X, we use the URL of X.  To find the probability of obtaining a score < X, we use the LRL of X.  To find the probability of obtaining a score = X, use the URL and LRL of X.

Sample Problem:  Suppose we conduct the following experiment on ESP. Five different cards are shuffled and one is chosen at random. Using ESP, the subject must try to guess the chosen card. This procedure is repeated for a total of 100 trials. If we assume that ESP doesn’t really exist, what is the probability that the student will be correct on more than 25 trials?

This is a binomial problem because the data in this example are binomial - on each trial the subject is correct or incorrect.  Can we legitimately use the normal approximation (are both pn & qn ³ 10)?

p = probability of a correct response = p(correct) = 1/5 = .20

q = probability of incorrect response = p(incorrect) = 4/5 = .80

n = 100

pn = .20(100) = 20 qn = .80(100) = 80 

Use real limits to express the probability question:  p(X > 25.5) = ??

Calculate m and s.

   m = pn  = .20(100) = 20

     s = square root of npq = sq root (100)(.2)(.8) = 4

Draw a normal distribution, label, and shade desired area.

Calculate z.  p(X > 25.5) = p(z > ??) = ??

       z = (X - m) / s = (25.5 - 20) / 4 = +1.375

Find the probability using the unit normal table.  Look up 1.375 in column A and find the corresponding value in column C (tail). 

                p(X > 25.5) = p(z > 1.375) = .0838

Answer the question:  It is very unlikely, .0838 or 8.38% probability, that you would obtain more than 25 correct responses if ESP does not exist.