Let x1, 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

answer( isp );