Answer:
The minimum percentage of noise level readings within 8 standard deviations of the mean is 98.44%.
Step-by-step explanation:
When we do not know the shape of the distribution, we use the Chebyshev's Theorem to find the minimum percentage of a measure within k standard deviations of the mean.
This percentage is:
[tex]p = 1 - \frac{1}{k^{2}}[/tex]
Within 8 standard deviations of the mean
This means that [tex]k = 8[/tex]. So
[tex]p = 1 - \frac{1}{8^{2}} = 1 - \frac{1}{64} = \frac{64 - 1}{64} = \frac{63}{64} = 0.9844[/tex]
The minimum percentage of noise level readings within 8 standard deviations of the mean is 98.44%.
