Step 1 :This problem is about the sampling distribution of proportion. The sampling distribution of the sample proportion is approximately normally distributed if the sample size is large enough. We can use the standard normal distribution (Z-distribution) to calculate the probability.
Step 2 :The formula to calculate the Z score is: \(Z = \frac{{p\_hat - p}}{{\sqrt{{\frac{{p*(1-p)}}{n}}}}\), where \(p\_hat\) is the sample proportion, \(p\) is the population proportion, and \(n\) is the sample size.
Step 3 :In this case, we want to find the probability that the sample proportion will differ from the population proportion by less than 0.04, which means we want to find \(P(-0.04 < p\_hat - p < 0.04)\). We can convert this to the Z score and find the corresponding probability from the Z-distribution.
Step 4 :Let's calculate the Z scores for the two boundaries (-0.04 and 0.04) and find the corresponding probabilities. The final answer will be the difference between the two probabilities.
Step 5 :Given that \(p = 0.05\) and \(n = 216\), we can calculate the standard deviation as \(\sqrt{{\frac{{p*(1-p)}}{n}}} = 0.014829275350043488\).
Step 6 :The Z scores for the two boundaries are calculated as \(Z1 = \frac{{-0.04 - p}}{{std\_dev}} = 0.6743417843388197\) and \(Z2 = \frac{{0.04 - p}}{{std\_dev}} = 6.069076059049377\).
Step 7 :The corresponding probabilities from the Z-distribution are \(prob1 = 0.7499529775707622\) and \(prob2 = 0.999999999356759\).
Step 8 :The final probability that the sample proportion will differ from the population proportion by less than 0.04 is \(prob = prob2 - prob1 = 0.2500470217859968\).
Step 9 :Rounding to four decimal places, the final answer is \(\boxed{0.2500}\).