Step 1 :The problem is asking to estimate the probability of getting fewer than 5 successes in a binomial distribution with parameters \(n=13\) and \(p=0.5\) using the normal approximation. The normal approximation to the binomial distribution is suitable when both \(np\) and \(nq\) are greater than or equal to 5. Here, \(np = 13*0.5 = 6.5\) and \(nq = 13*0.5 = 6.5\), so the normal approximation is suitable.
Step 2 :The mean of the binomial distribution is \(np = 6.5\) and the standard deviation is \(\sqrt{npq} = \sqrt{13*0.5*0.5} = 1.8028\).
Step 3 :We want to find \(P(X<5)\), but since we are using a continuous distribution to approximate a discrete one, we use the continuity correction and find \(P(X<4.5)\) instead.
Step 4 :To standardize, we subtract the mean and divide by the standard deviation, getting \(Z = \frac{4.5 - 6.5}{1.8028}\).
Step 5 :We can then look up this Z-score in the standard normal table to find the probability.
Step 6 :Final Answer: The probability of getting fewer than 5 successes is approximately \(\boxed{0.1336}\).