Example 4.3.3: Applying the Central Limit Theorem for proportions.
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A power tools manufacturer reviews the production history for all drill bits produced and found that $8 \%$ of the drill bits made are defective. A change in the manufacturer's quality assurance process reduced the percentage of defective drill bits to $5 \%$ of the 300 drill bits sampled. If the quality assurance process was not changed, what is the probability that at most $5 \%$ of drill bits are defective? The population proportion is $p=0.08$ and the sample proportion is $\hat{p}=0.05$.
Solution
The population proportion is $p=0.08$. Both $n p=300(0.08)=24$ and $n(1-p)=300(0.92)=276$ are greater than or equal to 5 . Thus, the conditions of the CLT are satisfied and the shape of the binomial distribution approaches that of the normal distribution. The standard deviation is
\[
\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.08(1-0.08)}{300}}=0.016
\]
The corresponding $z$-score is
\[
z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.05-0.08}{0.016}=-1.875
\]
Using the Excel function NORM.S.DIST(-1.875, TRUE) gives a probability of $\approx 0.0304$.
Answer
Final Answer: The probability that at most 5% of drill bits are defective if the quality assurance process was not changed is approximately \(\boxed{0.0277}\).
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 \(\boxed{0.0277}\).
