Final answer:
The minimum percentage of noise level readings within 2 standard deviations of the mean, without assuming a normal distribution, is calculated using Chebyshev's inequality. For 2 standard deviations (k=2), at least 75% of the values must fall within this range. The only choice that fits this criteria is D) 100%.
Step-by-step explanation:
The question asks for the minimum percentage of noise level readings within 2 standard deviations of the mean, without assuming a normal distribution. According to Chebyshev's inequality, which applies to any distribution whether it's normal or not, at least (1 - 1/k²) of the values lie within k standard deviations of the mean for k > 1. In this case, k equals 2, so the minimum percentage of noise levels within 2 standard deviations (59 dB to 115 dB) would be at least (1 - 1/2²), which simplifies to (1 - 1/4) or 0.75. Therefore, at least 75% of the noise levels should fall within 2 standard deviations of the mean.
Translated to percentage, this is at least 75.00%, which means the correct answer to the question is:
This response is however incorrect as the minimum percentage guaranteed by Chebyshev's inequality is lower than 95.45%. We should correct the selected choice to:
However, this response is also not the minimum percentage. The only choice that represents the minimum percentage guaranteed by Chebyshev's inequality is D) 100%. This is because we are only asked about the minimum percentage within the range, which implies at most 100% of the values can fall within two standard deviations of the mean.