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 5 dB. Without assuming a normal distribution, find the minimum percentage of noise level readings within 5 standard deviations of the mean During a rock concert, the noise level (in decibels) in front row seats has a mean of 95 dB with a standard deviation of 5 dB. Without assuming a normal distribution, find the minimum percentage of noise level readings within 5 standard deviations of the mean

Answer 1

We know about Chebyshev's theorem, we can find minimum percentage of noise level readings within 5 s.d. of the mean by using the following formula:-

`Percentage= 1 -(1)/(x^(2))`

Here the value of x=5 given in the question.

So
`1-(1)/(5^(2)) \;=\;1-(1)/(25)\;=\;(24)/(25) \;=\;0.96`

So final answer is 0.96 or 96%.