a= μ-3.16*σ , b= μ+3.16*σ if each fish caught is equally likely to be any one of these 4 distinct types.
<h3>What is meant by Chebyshev inequality?</h3>
Chebyshev's inequality is a probability theory that ensures that, over a vast range of probability distributions, no more than a particular proportion of values would be present within a selected limits or range as from mean. In other words, only a certain fish caught will be discovered within a given range of the distribution's mean.
The formula for which no more than a particular number of values can exceed is 1/K2; in other words, 1/K2 of a distribution's values can be more than or equal to K standard deviations away from the distribution's mean. Furthermore, it asserts that 1-(1/K2) of a distribution's values must be within, but not include, K standard deviations of the distribution's mean.
How to solve?
from Chebyshev's inequality for Y
P(| Y - μ|≤ k*σ ) ≥ 1-1/k²
where
Y = the number of fish that need be caught to obtain at least one of each type
μ = expected value of Y
σ = standard deviation of Y
P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean
k= parameter
thus for
P(| Y - μ|≤ k*σ ) ≥ 1-1/k²
P{a≤Y≤b} ≥ 0.90 → 1-1/k² = 0.90 → k = 3.16
then
P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90
using one-sided Chebyshev inequality (Cantelli's inequality)
P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)
P{Y≥b} ≥ 0.90 → 1- σ²/(σ²+λ²)= 1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ
then for
P(Y≥ μ+3*σ ) ≥ 0.90
In order to learn more about Chebyshev inequality, visit:
brainly.com/question/24971067
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