Question 1204322: The expected number of defective parts produced on an assembly line per shift is 50 with a standard deviation of 8. Use Chebyshev's inequality to find the minimum probability that the number of defective parts on a particular shift will be between 22 and 78. (Round your answer to four decimal places.)
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! **1. Define the Range**
* We are interested in the range of defective parts between 22 and 78.
* Mean (μ) = 50
* Standard Deviation (σ) = 8
**2. Calculate the Number of Standard Deviations (k)**
* **Lower Bound:** (50 - 22) / 8 = 3.5 standard deviations below the mean
* **Upper Bound:** (78 - 50) / 8 = 3.5 standard deviations above the mean
**3. Apply Chebyshev's Inequality**
* Chebyshev's Inequality states that for any data set, at least 1 - (1/k²) of the data values will fall within k standard deviations of the mean.
* In this case, k = 3.5
* Probability (22 ≤ Defective Parts ≤ 78) ≥ 1 - (1/3.5²)
≥ 1 - (1/12.25)
≥ 0.9184
**Therefore, according to Chebyshev's Inequality, the minimum probability that the number of defective parts on a particular shift will be between 22 and 78 is 0.9184.**
**Note:**
* Chebyshev's Inequality provides a lower bound for the probability. The actual probability may be higher.
* This inequality is applicable to any distribution, regardless of its shape.
Let me know if you have any other questions!
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