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Chapter 4: Problem 126

A quality-control plan calls for accepting a large lot of crankshaft bearings if a sample of seven is drawn and none are defective. What is the probability of accepting the lot if none in the lot are defective? If \(1 / 10\) are defective? If \(1 / 2\) are defective?

Short Answer

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Question: Calculate the probability of accepting the lot of crankshaft bearings in the given quality-control plan if: (1) none in the lot are defective, (2) \(1/10\) are defective, and (3) \(1/2\) are defective. Answer: - If none in the lot are defective: \(1\) - If \(1 / 10\) are defective: \(\approx 0.4783\) - If \(1 / 2\) are defective: \(\approx 0.0078\)

Step by step solution

01

Calculate the probability of accepting the lot if none are defective

Since none of the bearings in the lot are defective, the probability of drawing a non-defective bearing each time is \(1\). Therefore, the probability of drawing 7 non-defective bearings out of 7 is \(1^7 = 1\). So, the probability of accepting the lot if none are defective is \(1\).
02

Calculate the probability of accepting the lot if \(1 / 10\) are defective

If \(1 / 10\) are defective, then the probability of drawing a non-defective bearing is \(9 / 10\). Since we are drawing 7 bearings independently, we will find the probability of drawing all non-defective bearings using the binomial probability formula: \((9 / 10)^7\). Therefore, the probability of accepting the lot if \(1 / 10\) are defective is \((\frac{9}{10})^7 \approx 0.4783\).
03

Calculate the probability of accepting the lot if \(1 / 2\) are defective

If \(1 / 2\) are defective, then the probability of drawing a non-defective bearing is also \(1 / 2\). To find the probability of drawing 7 non-defective bearings out of 7, we use the binomial probability formula again: \((1 / 2)^7\). The probability of accepting the lot if \(1 / 2\) are defective is \((\frac{1}{2})^7 \approx 0.0078\). To summarize, the probability of accepting the lot: - If none in the lot are defective: \(1\) - If \(1 / 10\) are defective: \(\approx 0.4783\) - If \(1 / 2\) are defective: \(\approx 0.0078\)

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