Answer:
The minimum percentage of noise level readings within 3 standard deviations from the mean = 88.9%
Step-by-step explanation:
Since the type of distribution that this is, isn't known, we will use Chebyshev's formula to compute this percentage.
Chebyshev proposed that
P(|X - μ| ≥ kσ) ≥ [1 - (1/k²)]
This can be rewritten as
P(μ - kσ < X < μ + kσ) ≥ [1 - (1/k²)]
k must be greater than 1.
This is explained in words as, the probability of obtaining values from the data that lie within 'k-standard deviations' from the mean is at least [1 - (1/k²)]
So, for this question, to find the minimum percentage of noise level readings within 3 standard deviations of the Mean,
P(μ - kσ < X < μ + kσ) = [1 - (1/k²)]
(Equal to sign because we're looking for the minimum percentage.)
P(μ - 3σ < X < μ + 3σ) = [1 - (1/3²)]
[1 - (1/3²)] = 1 - (1/9) = (8/9) = 0.889 = 88.9%
Hope this Helps!!!
