Question
A manufacturing process produces components with a defect rate of \( 3% \). If you randomly select \( 400 \) components, what is the approximate probability that \( 9 \) or less components are defective?
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Solution
Answer: \( 23.2\% \)
We can model the number of defective components as \( X \sim B(400,0.03) \). The mean and standard deviation are:
\[ \mu = np = 400 * 0.03 = 12 \]
\[ \sigma = \sqrt{np(1-p)} = \sqrt{400 * 0.03 * 0.97} \approx 3.41 \]
Using a continuity correction for \( P(X \le 9) \), use 9.5:
\[ Z = \frac{9.5 - 12}{3.41} \approx -0.73 \]
Thus:
\[ P(X \le 9) \approx P(Z \le -0.73) \approx 23.2\% \]