Answer:
The procedures are below
Explanation:
Let X1,X2,...,X144 be independent and identically distributed random variables, each with expected value μ= E[Xi] = 2 and variance \sigma ^2= Var(Xi) = 4.
(a) Let Z=X1+ X2+...+X144, and use rules of Expectation and Variance to find E[Z]and Var[Z].
E(Z) = 144*E(xi) = 144*2 = 288
Var(Z) = 144*Var(Xi) = 144*4 = 576
sd (Z) = sqrt(576) = 24
(b) Let a be the difference between 144 and E[Z].
a = 144 - 288 = -144
(c) Apply Chebychev's Inequality to Z using the number a.
Statement of CHebyshev's inequality :
Let X (integrable) be a random variable with finite expected value μ and finite non-zero variance σ2. Then for any real number k > 0,
P(|X-mu| >=k*sigma) < = 1/k^2
Now we have to use this theorem for Z.
P(|Z-mu| >= k*24) < = 1/k^2
COmpare k*24 with 144
k*24 = 144
k = 144/24 = 6
P(|Z - 288| >= 144) <= 1/6^2
P(|Z - 288| >= 144) <= 0.0278
