Let X1, X2, ..., X144 be independent and identically distributed random variables, each with expected value? = E[Xi] = 2 and variance / sigma ^2 = Var (Xi) = 4. Find an upper bound for P (X1+X2+···+X144 >144) using the following steps:

(a) Let Z=X1 + X2 + ... + X144, and use rules of Expectation and Variance to find E[Z]and Var[Z].

(b) Let a be the difference between 144 and E[Z].

(c) Apply Chebychev's Inequality to Z using the number a.

(d) Use the fact that Z is symmetrically distributed about its mean to connect your answer to (c) to the original question. (Hint: Draw a symmetric density curve for Z, and mark the values E[Z], (E[Z]+a) and (E[Z]? a.)

Label regions in the graph with their corresponding probabilities.)

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