Final answer:
Using Chebyshev's inequality and knowing that 64 and 76 are 3 standard deviations from the mean of 70, we find that there's a minimum probability of 8/9, or 88.89%, that the number of defective parts will fall between 64 and 76.
Step-by-step explanation:
The student asked for the minimum probability that the number of defective parts produced on an assembly line per shift will be between 64 and 76. Given the expected number of defective parts (mean) is 70 with a standard deviation of 2, Chebyshev's inequality can be applied. To use Chebyshev's inequality, we need the number of standard deviations away from the mean that 64 and 76 are. In this case, 64 and 76 are 3 standard deviations away from the mean (70 ± 3(2)).
Chebyshev's inequality states that for any k>1, the probability that a random variable lies within k standard deviations (σ) of the mean (μ) is at least 1 - 1/k^2. Here, k is 3, so the inequality would be 1 - 1/3^2 = 1 - 1/9 = 8/9. Therefore, there's a minimum probability of 8/9, or approximately 88.89%, that the number of defective parts will be between 64 and 76.