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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 12 "A Fair Shake" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 9 "Pam Poppe" }}{PARA
263 "" 0 "" {TEXT -1 21 "ppoppe@mindspring.com" }}{PARA 262 "" 0 ""
{TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 21 "Brookwood High School"
}}{PARA 260 "" 0 "" {TEXT -1 30 "Gwinnett County Public Schools" }}
{PARA 261 "" 0 "" {TEXT -1 7 "Georgia" }}{PARA 256 "" 0 "" {TEXT -1 0
"" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "Problem Situation" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 206 "We simulate th
rowing a pair of fair dice with six faces for 360 times using a random
number generator and tabulating the sums of the top faces. To keep th
e die separate, we choose a red one and a blue one. " }}{PARA 0 "" 0 "
" {TEXT -1 218 " As an example of how to predict the three hundred
sixty outcomes, we predict how many times one would expect a sum of f
ive for the top faces of the two dice. First, recognize that a five ca
n be obtained four ways:" }}{PARA 0 "" 0 "" {TEXT -1 35 " \+
red 1 and blue 4" }}{PARA 0 "" 0 "" {TEXT -1 35 " \+
red 2 and blue 3" }}{PARA 0 "" 0 "" {TEXT -1 35 " \+
red 3 and blue 2" }}{PARA 0 "" 0 "" {TEXT -1 36 " \+
red 4 and blue 1." }}{PARA 0 "" 0 "" {TEXT -1 331 "Because throwing an
y single side has probability 1/6 and the throw of the red and the blu
e die are independent of each other, the probability of any of the cho
ices above would be 1/36. Because there are four choices, we surmise t
hat we will get a sum of five 4/36 = 1/9th of the time. In 360 throws,
this would be 360/9 = 40 times." }}{PARA 0 "" 0 "" {TEXT -1 146 " \+
In a similar manner, we can predicted the number of times each sum be
tween 2 and 12 occurs. The theoretical, predicted results are tabulate
d." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 134 " \+
Sums 2 3 4 5 6 \+
7 8 9 10 11 12" }}
{PARA 0 "" 0 "" {TEXT -1 125 " Predicted 10 20 3
0 40 50 60 50 40 30 \+
20 10" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 83 "For example, we recorded a predicted sum of five for 40 t
imes out of the 360 rolls." }}{PARA 0 "" 0 "" {TEXT -1 172 " We us
e the random number generator of Maple to simulate doing what no human
wants to do: We simulate throwing a pair of dice 360 times and keepin
g track of the result." }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 46 "Three Hund
red Sixty Virtual Throws of the Dice" }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 257 "" 0 "" {TEXT -1 113 "Before executing the first command c
hange the seed number to the last four digits of your social security \+
number." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "with(stats):with(
describe):_seed:=5064;" }}}{PARA 0 "" 0 "" {TEXT -1 83 "Each throw of \+
the red die or the blue die will be a random integer between 1 and 6.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "red:=rand(1..6):\nblue:=
rand(1..6):" }}}{PARA 0 "" 0 "" {TEXT -1 111 "Because we don't want to
count the results, we let Maple keep a count; first we zero counters \+
between 2 and 12." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "for i f
rom 2 to 12 do\n Count[i]:=0:\nod:" }}}{PARA 0 "" 0 "" {TEXT -1
70 "Here go the throws, counting the result of each throw of the two d
ice." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "for i from 1 to 360 \+
do \n n:=red()+blue():\n Count[n]:=1+Count[n]\nod:" }}}{PARA
0 "" 0 "" {TEXT -1 22 "Now we see the result." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 50 "for i from 2 to 12 do \n print(i,Count[i]) \+
\nod;" }}}{PARA 0 "" 0 "" {TEXT -1 166 "Toward making a histogram of t
he result, we take intervals around each of the integers two through t
welve, and weight the interval with the number of throws achieved." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "throws:=[seq(Weight(i-1/2...
i+1/2,Count[i]),i=2..12)];" }}}{PARA 0 "" 0 "" {TEXT -1 71 "We calcula
te the mean and standard deviation of the outcomes generated." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "mean(throws)=evalf(mean(thro
ws),3);\nstandarddeviation(throws)=evalf(standarddeviation(throws),3);
" }}}{PARA 0 "" 0 "" {TEXT -1 83 "We calculate the mean and standard d
eviation of the predicted theoretical outcomes." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 263 "theory:=[Weight(3/2..5/2,10),Weight(5/2..7/2,
20),Weight(7/2..9/2,30),\n Weight(9/2..11/2,40),Weight(11/2..13/2,
50),Weight(13/2..15/2,60),\n Weight(15/2..17/2,50),Weight(17/2..19
/2,40),Weight(19/2..21/2,30),\n Weight(21/2..23/2,20),Weight(23/2.
.25/2,10)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "mu:=mean(the
ory);\nsigma:=standarddeviation(theory);\nevalf(sigma,3);" }}}{PARA 0
"" 0 "" {TEXT -1 112 " We note that these theoretical computations
are close to the numbers obtained with the 360 virtual throws. " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 1 " " }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 23 "The Normal Distributio
n" }}{PARA 0 "" 0 "" {TEXT -1 549 " The relationship that assigns \+
to each number the probability that that number will be the sum of the
two faces is called a probability distribution function. The probabi
lity distribution function described here is a discrete probability fu
nction because it has a finite domain (the integers from 2 through 12)
. When there are infinitely many possible outcomes within some real n
umber interval the domain of the probability distribution function mig
ht be an interval. The probability function is called a continuous pr
obability distribution. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 308 "A histogram is a good graphic display of a discre
te probability. The bars of the histogram we present have the same wi
dth, and their heights are determined by the probabilities of the vari
ous values of the random variable. The area of each bar or rectangle \+
gives the probability for a particular outcome. " }}{PARA 0 "" 0 ""
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 203 "The problem situation a
bove approximates a normal distribution. The normal curve can be used
to approximate the probability of the sum of two six sided dice. The
equation for the normal distribution is" }}{PARA 0 "" 0 "" {TEXT -1
59 " " }
{XPPEDIT 18 0 "y=exp(-(x-mu)^2/(2*sigma^2))/(sigma*sqrt(2*Pi))" "6#/%
\"yG*&-%$expG6#,$*&,&%\"xG\"\"\"%#muG!\"\"\"\"#*&F0F-*$%&sigmaGF0F-F/F
/F-*&F3F--%%sqrtG6#*&F0F-%#PiGF-F-F/" }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 95 "The above equation is multiplied by 360 to account for \+
the number of times the dice are rolled." }}{PARA 0 "" 0 "" {TEXT -1
10 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "nd:=x->360
*exp(-(x-mu)^2/(2*sigma^2))/(sigma*sqrt(2*Pi));" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 83 "J:=statplots[histogram](theory):\nK:=plot([x,n
d(x),x=0..14]):\nplots[display](\{J,K\});" }}}{PARA 0 "" 0 "" {TEXT
-1 68 " We also compare the normal distribution with the virtual t
hrow." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "J:=statplots[histog
ram](throws):\nK:=plot([x,nd(x),x=0..14]):\nplots[display](\{J,K\});"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 41 "Probability from the No
rmal Distribution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 101 " Given, a continuous probability distribution functi
on f, the area under the curve f(x) from x = " }{TEXT 256 1 "a" }
{TEXT -1 8 " to x = " }{TEXT 257 1 "b" }{TEXT -1 51 " represents the p
robability that x will be between " }{TEXT 258 1 "a" }{TEXT -1 5 " and
" }{TEXT 259 1 "b" }{TEXT -1 183 ". The probability of an event happe
ning in the range from a to b can be computed from the normal distribu
tion we have associated with this experiment. This probability is an i
ntegral:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
27 " " }{XPPEDIT 18 0 "int(nd(x),x=a..b)" "6
#-%$intG6$-%#ndG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 1 "." }}{PARA 0 "" 0
"" {TEXT -1 193 "For example, let's compare the number of times the su
m of 7 occurred in the random generated data above with the theoretica
l predicted probability and with the integral of normal distribution. \+
" }}{PARA 0 "" 0 "" {TEXT -1 90 "The percentage of times the sum of 7
was generated out of the 360 generated tosses is ---" }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 24 "evalf((Count[7])/360,3);" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 72 " The theoretical probability of getting a
sum of 7 comes from the count " }}{PARA 0 "" 0 "" {TEXT -1 59 " 1 +
6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1." }}{PARA 0 "" 0
"" {TEXT -1 13 "This is 6/36." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "ev
alf(6/36,3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "Finding the area
under the normal curve from x = 6.5 to x = 7.5 is an approximation us
ing the normal distribution." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "In
t(exp(-(x-mu)^2/(2*sigma^2))/(sigma*sqrt(2*Pi)),x=6.5..7.5);\nint(exp(
-(x-mu)^2/(2*sigma^2))/(sigma*sqrt(2*Pi)),x=6.5..7.5);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1
8 "Exercise" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 115 "a) Simulate throwing a pair of dodecahedron dice (twelve side
d dice) 1440 times using a random number generator. " }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "b) Calculate the mea
n and standard deviation for your sample and compare them with the out
come probabilities. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 111 "c) Draw a histogram for the simulated throws on the \+
same graph as a continuous, normal distribution defined by" }}{PARA 0
"" 0 "" {TEXT -1 65 " \+
" }{XPPEDIT 18 0 "y=exp(-(x-mu)^2/(2*sigma^2))/(sigma*
sqrt(2*Pi))" "6#/%\"yG*&-%$expG6#,$*&,&%\"xG\"\"\"%#muG!\"\"\"\"#*&F0F
-*$%&sigmaGF0F-F/F/F-*&F3F--%%sqrtG6#*&F0F-%#PiGF-F-F/" }{TEXT -1 1 ",
" }}{PARA 0 "" 0 "" {TEXT -1 84 " where the mean and standard de
viation are the same as for the predicted sums." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161 "d) Compare the number \+
of times the sum of 13 occurred in the random generated data with the \+
predicted probability and the approximation from the normal curve. " }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 169 "Acknowle
dgement: The preparation of this worksheet was accomplished while the
author was a participant in Gift Summer Program at Georgia Tech durin
g the Summer of 1997." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }