Step 1 :The population proportion is \(p=0.08\). Both \(np=300(0.08)=24\) and \(n(1-p)=300(0.92)=276\) are greater than or equal to 5. Thus, the conditions of the Central Limit Theorem are satisfied and the shape of the binomial distribution approaches that of the normal distribution.
Step 2 :The standard deviation is calculated as \(\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.08(1-0.08)}{300}}=0.016\).
Step 3 :The corresponding z-score is calculated as \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.05-0.08}{0.016}=-1.875\).
Step 4 :Using the z-score, we can find the probability using a standard normal distribution table or a function like NORM.S.DIST in Excel. This gives a probability of approximately 0.0304.
Step 5 :Final Answer: The probability that at most 5% of drill bits are defective if the quality assurance process was not changed is approximately \(\boxed{0.0277}\).