Question

During a rock concert, the noise level (in decibels) in front row seats has a mean of 92 dB with a standard deviation of 9 dg_ Without
assuming a normal distribution, find the minimum percentage of noise level readings within 8 standard deviations of the mean. (Round
your answer to 2 decimal places.)
Minimum percentage

Answer 1

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:

$p = 1 - \frac{1}{k^{2}}$

Within 8 standard deviations of the mean

This means that $k = 8$. So

$p = 1 - \frac{1}{8^{2}} = 1 - \frac{1}{64} = \frac{64 - 1}{64} = \frac{63}{64} = 0.9844$

The minimum percentage of noise level readings within 8 standard deviations of the mean is 98.44%.