During a rock concert, the noise level (in decibels) in front row seats has a mean of 95 dB with a standard deviation of 8 dB. Without assuming a normal distribution, find the minimum percentage of noise level readings within 3 standard deviations of the mean. (Round your answer to 2 decimal places.)Minimum percentage%

Minimum percentage = 88.89%

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 95 dB

Standard Deviation, σ = 8 dB

Chebyshev's rule:

We have to find the minimum percent of data lying within 3 standard deviations of the mean.

Putting values, we get,

Thus, minimum 88.895 of data lies within 3 standard deviations of the mean.

17   x 256=4352

sorry if this is wrong, i'm new.

step-by-step explanation:

step-by-step explanation:

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about the problem:

srry i'm not done ! i'll