Question

1.3 Let X є {-1, +1} denote the outcome of an toss of an unbiased coin. (That is, Pr(X +1]-Pr[X-_1] = 1/2.) say the coin in tossed 1000 times independently, and the correspoinding outcomes are denoted by X1,... , X1000 Give a good estimate of the chance that the average of the 1000 tosses exceeds the value 101 That is, give the best possible value of a, such that Pr(x, + X1000) > 10] 〈 α.

Answer 1

S = X1 + X2+ ..X1000

E(S) = 1000E(X) = 1000 * 0 = 0

sd(S) = sqrt(Var(S))

Var(S) = 1000* Var(X)

= 1000 * 1 = 1000

sd(S) = sqrt(1000)

Z = (S - 0)/sqrt(1000)

P(S > 10)

= P(Z >(10 - 0)/sqrt(1000))

= P(Z > 1/sqrt(10))

= P(Z > 0.31622 )

= 0.3759

$\alpha$ can be any value > 0.3759

sat 0.38