Question

Define X1 = Z1, X2 = 22, ..., Xn = Zn and X = 36 L3fXi.

Consider the following probability A=P(x=11<3). (d) Please provide the distribution of X and find the exact probability A (a

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Answer #1

(d)

X1, X2,.., Xn are the squares of standard normal variables. Thus,

Xi follows Chi Square distribution with 1 degree of freedom.

E[Xi] = 1 and Var[Xi] = 2

By Central limit theorem,

\bar{X} ~ N(\mu = E[Xi] , \sigma^2 = Var[Xi] /36)

or,

\bar{X} ~ N(\mu = 1, \sigma^2 = 2/36)

\sigma = V2/36 = \sqrt{2} / 6

A = P(| 8 – 11<V2/3)=P(-V2/3<Ě -1<V2/3)

=P(-2< (7 - 1)/(V2/6) <2)

= P(-2<Z < 2)   (1 - 1)/(V2/6) ~ N(0,1)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544

(e)

By Chebyshev's inequality,

P(X – E[X]]>b) < Var(X)/64

Using b = \sqrt{2} /3

P(|X – 11 2 V2/3) < (2/36)/(V2/3)?

P(|X – 11 2 V2/3) < 0.25

1 - P(|X – 11 < V2/3) < 0.25

P(x - 1<V2/3) 21 – 0.25

p(\X – 11 <V2/3) > 0.75

Lower Bound for A is 0.75

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