Description
1 Two dice
Two unbiased dice are rolled. Let V be the sum of their face values.
a) Determine the probability with which each value of V occurs.
b) Determine the mean value V . Is this equal to two times the mean value for a single die?
c) Determine the standard deviation of V . Is this equal to two times the standard deviation
for a single die?
2 Multiple coin flips
This uses a computer simulation developed by the text authors; instructions for accessing
this were given in Homework 1. Run the “launcher” and select Multiple Coin Toss.
The simulation flips a set number of coins (e.g. 100) and records the number of heads. This
constitutes a single “trial” and the simulation can do repeated trials, record the results and
display these in a histogram. Each “trial” of the simulation uses many (e.g. 100) coin tosses.
The histogram plots the number trials that return various numbers of heads.
a) Set the probability to 0.5 and the number of coins to 100. In one run how many heads
do you predict to appear?
b) Click Initialize and then Step. How many flips produced heads? What does the data
yield for the average number of heads? Does this match your prediction?
c) Now repeat this many times by clicking Start and Stop until the simulation has done at
least 500 trials. Using the Tools → Data Tools menu in the Histogram window, determine
the most commonly occurring number of heads. What is the average number of heads
returned over all the trials? Does this match your prediction better than the result with
just one trial?
d) Now reset the simulation so that a single trial consists of 10000 flips and repeat steps a)
and b). Do the results of the simulation match your predictions more or less accurately
than when there were just 100 flips?
e) Repeat the previous part many times by clicking Start and Stop until the simulation
has done at least 100 trials. What is the average number of heads returned over all
the trials? Does this match your prediction better than the result with just one trial of
10000 flips? Does the histogram of the outcomes indicate a narrower or larger range of
possibilities than when there were just 100 flips?
f) What does the animation indicate about the accuracy of statistical predictions: do more
trials or fewer trials result in more accurate predictions?
1
3 Random walk
Consider a random walk with 6 steps.
a) If the probability of stepping right in a single step is 1/2, determine the probability with
which the walker ends two steps left of where he started.
b) If the probability of stepping right in a single step is 1/3, determine the probability with
which the walker ends two steps left of where he started.
4 Binomial distribution
Consider the binomial distribution whose outcomes are labeled +1 and −1. In a single trial
the probability of attaining +1 is p and that of attaining −1 is q. Suppose that N trials are
performed. The the probability with which one attains n outcomes of +1 is
P(n) = !N
n
”
pnqN−n.
a) Use the binomial theorem/identity to show that the probabilities satisfy
#
N
n=0
P(n)=1.
b) Use the fact that, for a binomial distribution
n2 = N2p2 + N pq
to determine the standard deviation σ for the distribution.
c) Consider the size of the standard deviation relative to the mean. If this is small, then
it means that the mean captures most of the information about the distribution. A
relevant measure of the size of the standard deviation to the mean is σ/n. Determine
an expression for this in terms of p and N. As N → ∞, what does this approach?
5 Binomial distribution: large numbers
This uses a computer simulation developed by the text authors; instructions for accessing
this were given in Homework 1. Run the “launcher” and select Binomial distribution. This
produces two graphs of probability distributions. One is P(n) versus n and the other is P(n)
versus n/ $n%. In this exercise you will use the graph of P(n) versus n. Also left clicking the
cursor on the graph gives the coordinates of any point.
Consider a collection of spin-1/2 particles, each of whose state can either be up or down. Let
N be the total number of particles and n the number whose spins are up. Suppose that p is
the probability with which any single particle has spin up.
a) Let N = 6 and suppose that p = 1/2. For each possible value of n, calculate the
probability P(n) with which n spins are up. Determine n and σ.
2
b) Let N = 10 and suppose that p = 1/2. Use the program to determine the probabilities
P(n) and to plot these. A measure of the width of the distribution is the Full Width
Half Maximum. One attains this by determining the maximum value of P(n) and
then determining the two values n1 and n2 so that P(n1) and P(n2) are half of this
maximum. The full width is then |n2 − n1|. Use the graph to estimate the FWHM for
the distribution.
c) Repeat this for N = 50, 200 and 1000. Does the FWHM increase with N? If so does it
scale linearly?
d) A crucial feature of the narrowness of such distributions is the value of the FHWM relative to n. Does the ratio FWHM/n increase, decrease or stay constant as N increases?
3
2020 HW 13 Q1
a) For a single che the probabilty of each number is /6
We consider all possibihties
Die Dc 2 Sun Prob V= Sum Prcb
2 1/36
2
/36
1 2 3 1136 3 2/36
2 3
4 3/36
3 4
5
2 2 4 all 1/36 4/36
2 4 6 5/36
4 5 7 6/36
2 3 S
8
5/36 3 2 S
4
5 9 4/36
10 3/36 ek
い 2/36 2 6
8
3 5 8 12 36
4 4 8
5 3 8
6 2 8
b) V= IV p(v) = 36[2x12x1 +3×2 +4×3 + 5×4+ 6×5 +7×6 +8×5
+9×4 +10×3 +11×2+ 12×1]
=7 For a single die the mear is 3.5
So yes this is twice
c) Need Ov= / 12 -)(2
+(2²x1+ 3²x2 +
Then vz = { v² plu) =
+ 12²x1) = 54.8 36
So ov= √54.8-72= 2.41 For a single dhe o =1.71 so not huice
Phys 362
2020 HW 13 QZ
) We expect
N heads prob= =D
N olal Nheads = prdb x N totul
= 0.5×100 =50
6) The simulation gave 46 heads
<H)= 46
Does not match exactly
C) Simulation gives: (504 tnals)
Mest commonly occwing : 50.5
average 50.246
It matches much better
d) Again precdict 5000
we get 5010
Much Mare accrate
e) Did 104 trials.
then 100 fups
<н) = 5005.683
Slightly better than one trial
Narrower histogram
f) Mare trials =o mare accrate.
(Means 50)
Let N = total nunber steps
NL= number left
Ne= nght and N =NL+NR
Then position is X= NR-NL
Ne-(N-Ne)
= 2NR-N
2020 HW 13 Q3
a) We need X=-2 =D -2 = ZNe-6
=0 4= 2N2 =D Ne=2
So we need probabilily that there are wo steps night. The
probabilly of k steps right is
P(lk)= ()pkаNk.
Here p=q=1/2
IN-K
=
2N
6! 6×5 15
26 412! 26 2 26
P(2)= 15 = 0.23
64
b) P(k) = (c)pk q N-k
P=3
q= 1-p = 2/3
4
P(*)= ()()*()
こ 15
T
36
2020 HW 13 Q3
24 246 80
그 =0.33
9x9x9 243
2020 HW 13 Q4
a) Σ P(n) = () Pa
こ
But the binomial thearem giues
N
(Pig) N= Σ ) p²q N-k.
and cleauly
K0
(P+q) N= Pa)
n=0
But prq =1 hee and thus
>P(n)= .
=0
b) G” < n²02-2
We have seen that
ñ = Np
and
² = ² + pqN
Thus. n² -ñ²= pq N
=D σ= √pqN= VP(1-p) VN
= Vpl-P) √N
Np
V
As N-O 0 this approaches zero.
a) P(n) = () P^ )-p)N-V
6-0
U
() 26
(5)
0 /64
6 6/64
2 15 15/64
3 20 20/64
4
15 15/64
5
6 664
6 64
2020 HW 13 Q5
61 30
= =15 2 42 2
6! 6.5.4 -20
3!3! 6
3 x1 +1×6 + 2×15 +3×20 + 4×15+ 5×6+6×1]
192 =6+30+60 +60 +3046] = 64 = 3 = 3
From previma problem nt =ñ²+ Np(lp) =o n² -^²= Np(-p)
b) max (at n=5) is 0.246
half max is 0.123
range for half max is 3.1-6.9
FWHM = 3.8
N range FWHM
FWHM لا
50 20.6- 29.4 8.8 25 0.352
200 91.5- 108.5 19 100 0.19
1000 481-518 37 500 0.074
It does increase, but not linearly
d) ving data above we see F WHM decreaes as N increases
