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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

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aachen

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:-

[tex] Percentage= 1 -\frac{1}{x^{2}} [/tex]

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

So [tex] 1-\frac{1}{5^{2}} \;=\;1-\frac{1}{25}\;=\;\frac{24}{25} \;=\;0.96 [/tex]

So final answer is 0.96 or 96%.

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