Approximating the Binomial with the Normal

Normal Approximation to Binomial:

Remember that according to the Central Limit Theorem, the sample mean of distribution will become around normal when the sample size is sufficiently big.

The binomial distribution can be approximated by using the normal distribution if both np and nq are at least 5. Moreover, remember that the mean of binomial distribution is np and the variance of binomial distribution is npq.

Continuity Correction Factor:

Approximating the binomial with normal is a big problem. That problem occurs as the binomial distribution is a discrete distribution whereas the normal distribution is a continuous distribution. The fundamental difference here is that with discrete values, we are talking regarding heights although no widths, and with continuous distribution we are talking regarding both widths and heights.

The correction is to either subtract or add 0.5 of a unit from each discrete x-value. This fills in gaps to make it continuous. This is much similar to expanding limits to form boundaries which are done with group frequency distributions.

Illustrations:

Discrete     Continuous
x = 6          5.5 < x < 6.5
x > 6          x > 6.5
x >= 6        x > 5.5
x < 6          x < 5.5
x <= 6        x < 6.5

As we can see, whether or not equal to be included makes a large difference in the discrete distribution and the manner the conversion is executed. Though, for a continuous distribution, equality makes no difference.

For binomial distribution, the steps for working a normal approximation are as follows:

A) Recognize success, probability of success, number of trials, and the desired number of successes. As this is a binomial problem, these are similar things that were recognized whenever working a binomial problem.

B) Transform the discrete x to a continuous x. Some of the people would argue that step 3 must be executed before this step, however go ahead and transform the x before you forget about it and fail to notice the problem.

C) Determine the smaller of np or nq. When the smaller one is at least five, then the bigger should also be, therefore the approximation will be considered good. Whenever you find np, you are really finding the mean, mu, so symbolize it as such.

D) Determine the standard deviation, sigma = sqrt (npq). It may be simpler to find the variance and just stick to the square root in the final calculation - that way you do not have to work with all decimal places.

E) Calculate the z-score by using the standard formula for an individual score (that is, not the one for sample mean).

F) Compute the probability desired.

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