Assume you have a fair coin and wish to know the probability that you would get eight heads out of ten flips. The binomial distribution has a mean of μ = np = 10 * 0.5 = 5 and a variance of σ² = npq = 10 * 0.5 * 0.5 = 2.5. The standard deviation is therefore 1.5811. A total of 8 heads is 1.8973 standard deviations above the mean of the distribution [(8–5)/1.5811]. The question then is ‘What is the probability of getting a value exactly 1.8973 standard deviations above the mean?’. The answer to this question is to remember that the probability of a particular event for a normal distribution is zero given that a particular event (or value of X) will not have an actual area within the normal distribution. The problem is that the binomial distribution is a discrete probability distribution whereas the normal distribution is a continuous distribution. The solution is to round off and consider any value from 7.5 to 8.5 to represent an outcome of 8 heads.

Using this approach, we can solve discrete binomial problems with a normal approximation if we transform X = 8 for the binomial to the region 7.5–8.5 for the normal distribution.

The area shaded in Figure 4.38 is an approximation of the probability of obtaining eight heads

We can see that the binomial probability distribution solution, P(X = 8) _Binomial ≈ P (7.5 ≤ X ≤ 8.5) _Normal.

Transcribed Image Text:

Normal curve f(x) T 5 6 7 8 P(X= 8) = 0.043495 T 9 X Figure 4.38