Answer:
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
