Uring a rock concert, the noise level (in decibels) in front row seats has a mean of 95 dB with a standard deviation of 8 dB. Without assuming a normal distribution, find the minimum percentage of noise level readings within 3 standard dev

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!!!

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

answer: 34rtyj,jkhtyetwrhyjeuiejnbhdfknbkguxfrbhkfdbjvkmbd,kvbsd.kxjvbhjnfbgjxdfn

step-by-step explanation: