Question 1204410: 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 ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Define Variables**
* Let X be the number of defective parts produced in a shift.
* Mean (μ) = 50
* Standard Deviation (σ) = 8
**2. Determine the Number of Standard Deviations**
* Calculate the distance from the mean to the lower and upper bounds:
* Distance from mean to lower bound (22): 50 - 22 = 28
* Distance from mean to upper bound (78): 78 - 50 = 28
* Calculate the number of standard deviations (k):
* k = Distance from mean / Standard deviation = 28 / 8 = 3.5
**3. Apply Chebyshev's Inequality**
* Chebyshev's Inequality states:
* P(|X - μ| ≥ kσ) ≤ 1/k²
* This means that the probability of the random variable X deviating from the mean by more than 'k' standard deviations is at most 1/k².
* In this case:
* P(|X - 50| ≥ 3.5 * 8) ≤ 1 / 3.5²
* P(|X - 50| ≥ 28) ≤ 1 / 12.25
* P(|X - 50| ≥ 28) ≤ 0.0816
* This inequality tells us that the probability of the number of defective parts being outside the range of 22 to 78 (i.e., deviating from the mean by more than 3.5 standard deviations) is at most 0.0816.
**4. Find the Minimum Probability within the Range**
* To find the minimum probability of the number of defective parts being between 22 and 78, we subtract the upper bound from 1:
* Minimum probability = 1 - 0.0816 = 0.9184
**Therefore, 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 general bound. If the distribution of the number of defective parts were known (e.g., Poisson distribution), a more precise estimate could be obtained.
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