Solutions for Quiz 4
1)
Suppose that 200 fair dice are tossed. Estimate the chance that the sum of the 200 rolls is at least
715.
a)
Box Model:
Draw from the following box and sum draws:
[1, 2, 3, 4, 5, 6]
Box Statistics:
Box.avg = (1+2+3+4+5+6)/6 = 3.5
Box.sd = 1.707825
Sum of draws Statistics:
1) EV = 200*3.5 = 700
2) SE = sqrt(200)*1.707825 = 24.152295
Min = 200
Max = 1200
b)
Normal Approximation:
1)
Continuity Correction: Necessary since we have at least one pair of consecutive
integers, endpoint is included, and # of draws is less than 10,000.
2)
Convert 715.5 = 714.5 to Standard Units,
SU = (714.5 – EV)/SE = (714.5 – 700)/24.152295 = .600357
Area above this is,
50% – 0.5*Area(z=.600357) = 50% – 0.5*45.15% = 27.43
c)
Normal Approximation Justification:
1)
There are at least 25 draws
2)
700 – 2*SE > 200 and 700 + 2*SE < 1200
2)
a)
Box Model: Draw from the following box:
[
From each pair of bets your net gain is either $2 (Lose, Lose), $0 (Win, Lose), $34
(Lose,
Win), or $36 (Win, Win).
The chance of getting $2 is
(20/38)*(37/38)= 740/1444
The chance of getting $0 is (18/30)*(37/38) = 666/1444
The chance of getting $34 is (20/38)*(1/38) = 20/1444
The chance of getting $36 is (18/38)*(1/38) = 18/1444
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 anderes
 Statistics, Normal Distribution, Variance, Probability theory, Binomial distribution, probability density function, Normal approximation

Click to edit the document details