Question

In playing a game, you win or lose 1 dollar with probability 0.5. If you play the game independently 1,000 times, what is (approximately) the probability that your fortune (the total amount you won or lost) is at least 10 dollars? (Use the Central Limit Theorem)

Use Central Limit Theorem Please!

Answer 1

for single game:

x	P(x)	xP(x)	x2P(x)
1	0.500	0.500	0.500
-1	0.500	-0.500	0.500
	total	0.000	1.000
	E(x) =μ=	ΣxP(x) =	0.0000
	E(x2) =	Σx2P(x) =	1.0000
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.000
	std deviation=	σ= √σ2 =	1.0000

hence for 1000 game; expected gain/loss =0*1000 =0

and std deviaiton =1*sqrt(1000)=31.623

for normal distribution z score =(X-μ)/σx
here mean=       μ=	0
std deviation   =σ=	31.623

P(your fortune is at least 10 dollars):

probability =

P(X>10)

=

P(Z>0.32)=

1-P(Z<0.32)=

1-0.6255=

0.3745