Question

A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y denote the number of fish that need be caught to obtain at least one of each type.
(a) Give an interval (a, b) such that (b) Using the one-sided Chebyshev inequality, how many fish need we plan on catching so as to be at least 90 percent certain of obtaining at least one of each type?

Answer 1

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.

What is meant by Chebyshev inequality?

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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