Question

900 draws will be made at random with replacement from the box [2 4 6 8]. Estimate the chance that the sum of the draws will be more than 4,600. (Round two decimals)

Answer 1

x	P(X=x)	xP(x)	x2P(x)
2	0.250	0.50000	1.00000
4	0.250	1.00000	4.00000
6	0.250	1.50000	9.00000
8	0.250	2.00000	16.00000
total		5.0000	30.0000
	E(x) =μ=	ΣxP(x) =	5.0000
	E(x2) =	Σx2P(x) =	30.0000
	Var(x)=σ2 =	E(x2)-(E(x))2=	5.0000
	std deviation=	σ= √σ2 =	2.2361

expected sum of 900 draws =900*5 =4500

and standard deviation =2.2361*√900 =67.082

from normal approximation:

chance that the sum of the draws will be more than 4,600 :

P(X>4600)=P(Z>(4600-4500)/67.082)=P(Z>1.49)=0.07