Question

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.)

Answer 1

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

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
`1-(1)/(k^(2) )` where, k > 1.

So, according to our question, k = 3

Minimum percentage of noise level readings within 3 standard deviations of the mean =
`(1-(1)/(3^(2) ))*100` =
`(1-(1)/(9 ))*100` =
`(8)/(9) *100` = 88.89% .

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

Answer 2

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

Explanation:

We are given the following in the question:

Mean = 95 dB

Standard Deviation = 8 dB

Chebyshev's Rule:

• Atleast
  `1-(1)/(k^2)` percent of data lies within k standard deviation of mean for a non-normal data.
• For k = 3

`1-(1)/((3)^2) = 88.89\%`

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