Correction for Continuity

Since the normal distribution is a continuous distribution, the probability of getting any single value such is nil. Therefore, the normal approximation of the probability of getting exactly 5 successes is the area between 4.5 and 5.5. To see this, enter 5 in both the "from" and the "to" fields and hit return.

Consider the problem of estimating the probability of 7 or more successes out of 8 when the probability of success on any one trial is 0.6. Enter 8 for N, .6 for p, 7 for "from," 8 for "to," and hit the enter key. The binomial probability is 0.1064 and the normal approximation is 0.1099. Notice when the highest possible value is entered (8 in this case) the shaded area extends to the end of the distribution.

A first try at calculating the normal approximation
A mechanical application of the normal distribution to this problem would be as follows:

A normal distribution with paramters N = 8 and p = 0.6 has a mean of Np = 4.8 and a standard deviaiton equal to = = 1.386. The probability of 7 or more can then be calculated using the formula:

Z = (X-M)/sd = (7.0 - 4.8)/1.386 = 1.587

where M is the mean and sd is the standard deviation.

The probability of a Z greater than or equal to 1.587 is 0.0563 which is much lower than the binomial probability of 0.1064.

A better approximation
The problem with the first approach is that the area between 6.5 and 7.0 should be included in the probability of 7 successes. This can be accomplished by subtracting 0.5 from X before subtracting the mean from X. Therefore, the formula is:

Z = (X - 0.5 - M )/sd = (7.0 - 0.5 - 4.8)/1.386 = 1.227. The probability of a Z greater than or equal to 1.227 is 0.1099, the same value computed by the Java applet.