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 iterval (a, b) such that P{a≤Y≤b} ≥ 0.90b. 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.

a) P (μ-k*σ≤ Y ≤ μ+k*σ) ≥ 0.90

a = μ-3.16*σ, b = μ+3.16*σ

b) P (Y≥ μ+3*σ) ≥ 0.90

b = μ+3*σ

Step-by-step explanation:

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