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 88.89% percentage of noise level readings within 3 standard deviations of the mean.

Step-by-step explanation:

We are given the following in the question:

Mean = 95 dB

Standard Deviation = 8 dB

Chebyshev's Rule:

Thus, minimum 88.89% percentage of noise level readings within 3 standard deviations of the mean.

The minimum percentage of noise level readings within 3 standard deviations of the mean = 88.89% .

Step-by-step explanation:

We are given in the question that the noise level (in decibels) in front row seats has a mean of 95 dB with a standard deviation of 8 dB.

Also, it is not stated that data is normally distributed.

So, according to Chebyshev's theorem it has been stated that percentage of the data that lies within k standard deviations of the mean equals at least   where, k > 1.

So, according to our question, k = 3

Minimum percentage of noise level readings within 3 standard deviations of the mean = = = = 88.89% .

Hence, 88.89% is the minimum percentage of noise level readings within 3 standard deviations of the mean.

x^5/3y^1/3 the correct answer

hope it

8p+(-9)

(-9)3p+5p and you can swith em around