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Suppose all items from an assembly line are independent and that

the probability of any item being defective is 0.1 for all items.
If you sample a bunch of items and the standard deviation for the
number

1 Answer

3 votes

Final answer:

X represents the number of defective cars sampled from an assembly line with a 10% defect rate. With a sample size of 100, the mean is 10 and the standard deviation is 3, allowing us to apply the 68-95-99.7 empirical rule. Therefore, 68% of samples will have 7-13 defective cars, 95% will have 4-16, and 99.7% will have 1-19 defective cars.

Step-by-step explanation:

If we consider a NUMMI assembly line with a 10 percent defect rate, and we sample n = 100 cars, we can define the random variable X to represent the number of defective cars in the sample. Since the probability of a car being defective is 0.1, the expected number of defective cars (the mean, μ) in our sample is 0.1 * 100 = 10. The variance of X in a binomial distribution is given by σ2 = n * p * (1 - p), which for our case is 100 * 0.1 * (1 - 0.1) = 9. Hence, the standard deviation (σ) is the square root of the variance, σ = √9, which is 3.

Applying the 68-95-99.7 empirical rule (also known as the normal distribution rule), we can say that approximately:

  • 68% of the samples will have between 10 - 3 and 10 + 3 defective cars (7 to 13 defective cars).
  • 95% of the samples will have between 10 - 2*3 and 10 + 2*3 defective cars (4 to 16 defective cars).
  • 99.7% of the samples will have between 10 - 3*3 and 10 + 3*3 defective cars (1 to 19 defective cars).

This analysis assumes that the distribution of defective cars is approximately normal, which is justified by the large sample size according to the central limit theorem.

User Marybeth
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