Respuesta :
Answer:
Minimum 88.89% percentage of noise level readings within 3 standard deviations of the mean.
Step-by-step explanation:
We are given the following in the question:
Mean = 95 dB
Standard Deviation = 8 dB
Chebyshev's Rule:
- Atleast [tex]1-\dfrac{1}{k^2}[/tex] percent of data lies within k standard deviation of mean for a non-normal data.
- For k = 3
[tex]1-\dfrac{1}{(3)^2} = 88.89\%[/tex]
Thus, minimum 88.89% percentage of noise level readings within 3 standard deviations of the mean.
Answer:
The minimum percentage of noise level readings within 3 standard deviations of the mean = 88.89% .
Step-by-step explanation:
We are given in the question that the noise level (in decibels) in front row seats has a mean of 95 dB with a standard deviation of 8 dB.
Also, it is not stated that data is normally distributed.
So, according to Chebyshev's theorem it has been stated that percentage of the data that lies within k standard deviations of the mean equals at least [tex]1-\frac{1}{k^{2} }[/tex] where, k > 1.
So, according to our question, k = 3
Minimum percentage of noise level readings within 3 standard deviations of the mean = [tex](1-\frac{1}{3^{2} })*100[/tex] = [tex](1-\frac{1}{9 })*100[/tex] = [tex]\frac{8}{9} *100[/tex] = 88.89% .
Hence, 88.89% is the minimum percentage of noise level readings within 3 standard deviations of the mean.
