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.)Minimum percentage____%

Answer 1

Answer:

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:

• According to this rule atleast $1 - \dfrac{1}{k^2}$ percent of data lies within k standard deviation of mean.

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

Putting values, we get,

$1 - \dfrac{1}{(3)^2} = 88.89\%$

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