Final answer:
Using the empirical rule and standard deviation, the probability of a specimen having an acceptable hardness within 67 to 71 is likely 68.27%. For 96% acceptable hardness, ±c corresponds to a z-score of 1.96. The probability of at most 4 of 6 specimens being acceptable requires additional binomial distribution calculations.
Step-by-step explanation:
The probability that a randomly chosen specimen has an acceptable hardness within the range of 67 to 71 is best represented using standard deviation and z-scores. Since no other specific distribution information is provided, we assume a normal distribution. The value within a certain range in a normal distribution relates to the empirical rule or can be found using a z-score table. Assuming 69 is the mean, the range of hardness from 67 to 71 represents ±2 standard deviations. From the empirical rule, approximately 68% of all values lie within this range. Therefore, the answer to question a would likely be 0.6826894921.
For question b, the value of c that results in 96% of all specimens having acceptable hardness within the range (69-c, 69+c) is related to the concept of z-scores and the corresponding areas under the normal distribution curve. In a standard normal distribution, 96% corresponds to a z-score range of approximately ±2, which equates to 1.96.
Finally, for question c, if the acceptable range is defined in part (a), the probability that at most 4 of 6 specimens have acceptable hardness can be calculated using the binomial distribution formula or by looking up a binomial distribution table. However, this calculation requires additional information such as the exact probability of a single specimen having acceptable hardness or a distribution chart to reference.