Statistics and Analysis of Scientific Data (Instructor's Solution Manual) (Solutions)

87 Pages·2016·0.591 MB·English

by  Massimiliano Bonamente

Solutions Manual
Statistics and Analysis of
Scientific Data, second edition
Massimiliano Bonamente
December 22, 2016
Springer

Chapter 1
1.1.Describe the sample space of the experiment consisting of flipping four
coins simultaneously. Assign the probability to the event consisting of “two
heads up and two tails up”. In this experiment it is irrelevant to know which
specific coin shows heads up or tails up.
Solution:Thesamplespacefortheexperimentinwhich4coinsareflipped
insequence,andforwhichtheorderoftossisconsideredimportant,contains
24 events,startingfromtheeventinwhichall4coinsshowheadsup,HHHH,
toHHHTinwhichthelastshowstailsup,untilTTTT,inwhichallfourtosses
resulted in tails up. The list of all ordered outcomes is shown in Table1.1.
Each of these 16 events is assigned the same probability of 1/16, since we
assume that the coins are not biased.
The sample space for the experiment in which the order of toss is consid-
eredunimportantconsistsofjust5outcomes:4headsup,3headsand1tails
up, 2 heads and 2 tails up, 3 heads and 1 tails up, and all 4 tails up. Each of
these 5 events is the union of a number of the events in the experiment that
takes into account the order of the toss. By direct count of those events, we
find that the event consisting of any two heads up, and two tails up, has a
probability of 6/16, or 3/8.
Table 1.1 Listoforderedeventsforthe4-coinflipexperiment
HHHH HTHH THHH TTHH
HHHT HTHT THHT TTHT
HHTH HTTH THTH TTTH
HHTT HTTT THTT TTTT
1.2.An experiment consists of rolling two dice simultaneously and indepen-
dently of one another. Find the probability of the event consisting of having
either an odd number in the first roll or a total of 9 in both rolls.
3
4 1
Solution: For the event that consists of an odd number in the first roll
(A), the probability can be calculated independently of the outcome of the
second roll, since the two rolls are independent. The probability is therefore
P(A) = 1/2. A total of 9 (event B) is obtained in just 4 of the 36 possible
outcomes of the roll of two dice (3 and 6, 4 and 5, 5 and 4, and 6 and 3), for
a probability of P(B)=1/9.
The probability of occurrence of either event, A∪B, must exclude the
intersection of the two events to avoid double counting, which contains all
eventsinAthatarealsoinB,specificallytherollsequencesof3and6,5and
4, for a probability of the intersection of 1/18. The probability is therefore
P(A∪B)=1/2+1/9−1/18=5/9.
1.3.In the roll of a die, find the probability of the event consisting of having
either an even number or a number greater than 4.
Solution: The event A, consisting of an even number, has a probability
of 1/2, and the event B consisting of a number greater than 4 (either 5 or 6)
has a probability of 1/3. The intersection of these two events is the outcome
6, with probability 1/6, and therefore the union A ∪ B has a probability
P(A∪B)=1/2+1/3−1/6=2/3.
1.4.An experiment consists of rolling two dice simultaneously and indepen-
dently of one another. Show that the two events, “the sum of the two rolls is
8” and “the first roll shows 5” are not statistically independent.
Solution: The event consisting of a sum of 8 in two rolls (event A) has a
probabilityofP(A)=5/36(thereare5orderedcombinationsgivingasumof
8), and the event resulting in a 5 in the first roll (event B) has a probability
of P(B)=1/6.
Toprovethatthesetwoeventsarenot independent,wemustshowthatthe
product of their probabilities does not equal the probability of A∩B, which
consists of just one possible outcome (5 in first roll, and 3 in the second), for
P(A∩B)=1/36. Since P(A)P(B)=5/36×1/6(cid:54)=1/36, the two events are
not independent. The dependence between the two events is obvious, since a
5 in the first roll limits the possibilities of obtaining 8 as the total.
1.5.An experiment consists of rolling two dice simultaneously and indepen-
dently of one another. Show that the two events, “first roll is even” and
“second roll is even” are statistically independent.
Solution: An even number in the first roll has P(A) =1/2, and an even
number in the second roll also has a probability of P(B) = 1/2, since each
eventisconcernedjustwiththeoutcomeofonerollatatime.Theintersection
of the two events is given by the 9 sequences of all-even numbers, namely 2
1 5
and2,2and4,2and6;4and2,4and4,4and6;and6and2,6and4,and6
and 6, for P(A∩B)=1/4. Since P(A∩B)=P(A)P(B), the two events are
shown to be independent. The independence is obvious, since the two tosses
are carried out separately.
1.6.A box contains 5 balls, of which 3 are red and 2 are blue. Calculate (a)
the probability of drawing two consecutive red balls and (b) the probability
of drawing two consecutive red balls, given that the first draw is known to
be a red ball. Assume that after each draw the ball is replaced in the box.
Solution: The probability of drawing one red ball is p=3/5, since 3 of 5
balls are red.
(a) The consecutive draws are independent, and therefore the probability
oftwoconsecutiveredsis(3/5)2,sincethedrawsaremadewithreplacement.
(b)Theconditionthatthefirstdrawwasaredisirrelevanttotheoutcome
of the second draw, and therefore the probability is just 3/5. The solution
can be obtained also using the definition of conditional probability,
P(A∩B)
P(B/A)=
P(A)
where A is the event of a red ball in the first draw and B is the event of a
red ball in the second draw, with P(A∩B)=(3/5)2 and P(A)=3/5.
1.7.A box contains 10 balls that can be either red or blue. Of the first three
draws,donewithreplacement,tworesultinthedrawofaredball.Calculate
the ratio of the probability that there are 2 or just 1 red ball in the box and
the ratio of probability that there are 5 or 1 red balls.
Solution: The ratio between the presence of 2 and 1 red balls, given the
measurement B, is given by
P(A /B) P(B/A )P(A )
r = 1 = 1 1 ,
12 P(A /B) P(B/A )P(A )
2 2 2
where P(A ) and P(A ) are prior probabilities. Since the container a priori
1 2
may contain between 0 and 10 red balls with same probability, for a total
of 11 choices, we set P(A ) = 1/11 and P(A ) = 1/11, i.e., a uniform prior
1 2
distribution. The ratio is a convenient number to calculate, since it does not
require P(B), which is numerically more intensive to calculate:
(cid:0)3(cid:1)(cid:18) 1 (cid:19)2(cid:18)N −1(cid:19)×P(A )
P(A /B) 2 N N 1 (cid:18)1(cid:19)2 9
r = 1 = = × =0.28
12 P(A2/B) (cid:18) 2 (cid:19)2(cid:18)N −2(cid:19) 2 8
(3) ×P(A )
2 N N 2
6 1
where N =10. This was expected, since it appears considerably more likely
that there are two red balls, given that in three draws, twice a red ball was
drawn; since after the draw the ball is placed again in the container, it is
stillpossiblethat thereisjust oneredball. Usingthesame logic,the ratio of
posterior probabilities of i=1 and i=5 is given by:
(cid:0)3(cid:1)(cid:18) 1 (cid:19)2(cid:18)N −1(cid:19)×P(A )
P(A /B) 2 N N 1 (cid:18)1(cid:19)2 9
r = 1 = = × =0.072,
15 P(A5/B) (cid:18) 5 (cid:19)2(cid:18)N −5(cid:19) 5 5
(3) ×P(A )
2 N N 5
where we set P(A ) = 1/11. The result shows that it is considerably more
5
likely that there are 5 red balls rather than just 1 ball.
1.8.In the game of baseball a player at bat either reaches base or is retired.
Consider 3 baseball players: player A was at bat 200 times and reached base
0.310 of times; player B was at bat 250 times, with an on-base percentage of
0.296;playerCwasatbat300times,withanon-basepercentage0.260.Find
(a)theprobabilitythatwheneitherplayerA,BorCwereatbat,hereached
base, (b) the probability that, given that a player reached base, it was A, B
or C.
Solution:PlayerAhasanon-basepercentageof0.310in200tries,player
B of 0.296 in 250, and C of 0.260 in 300. For part (a), the probability that
when either player A, B or C were at bat, the player reached base, is given
by P(base/A) = 0.31, P(base/B) = 0.296, and P(base/C) = 0.26. Also, the
probability of each player being at bat was

P(A at bat)≡P(A)=200/750=0.267

P(B)=250/750=0.333
P(C)=300/750=0.400.
The Total Probability theorem can be used to find that the probability of
reaching base when any player was at bat is
P(base)=P(base/A)P(A)+P(base/B)P(B)+P(base/C)P(C)
=0.31×0.267+0.296×0.333+0.26×0.400=0.285.
For part (b), we can use Bayes’ theorem to find that

P(A/base)=P(base/A)P(A)/P(base)=0.31×0.267/0.285=0.290

P(B/base)=0.346
P(C/base)=0.365.
Thesumofthesethreeprobabilitiesequals1,towithintheroundofferrors.
1 7
1.9.An experiment consists of rolling two dice simultaneously and indepen-
dentlyofoneanother.Calculate(a)theprobabilityofthefirstrollbeinga1,
given that the sum of both rolls was 5, (b) the probability of the sum being
5, given that the first roll was a 1 and (c) the probability of the first roll
being a 1 and the sum being 5. Finally, (d) verify your results with Bayes’
theorem.
Solution: The probability of rolling a 1 (event A) is 1/6, and the proba-
bility of the sum being 5 (event B) is 4/36, since the sequences that give a
sum of 5 are just 4.
(a) The probability P(A/B) = 1/4, since there is just one of 4 possible
sequences with a 1 in the first roll.
(b) P(B/A) = 1/6, since having rolled a 1, there is only 1 in six
combinations—namely a 4 in the second roll—that gives a sum of 5.
(c)ItisalsoclearthatP(A∩B)=1/36,sincethereisjust1in36possible
sequences to satisfy both requirements.
1 4 1
P(A∩B)=P(A/B)×P(B)= ×( )=
4 36 36
(d) You can verify that
P(B/A)P(A) 1/6×1/6 1
P(A/B)= = =
P(B) 4/36 4
to show that Bayes’ theorem is verified.
1.10.Four coins labeled 1 through 4 are tossed simultaneously and indepen-
dently of one another. Calculate (a) the probability of having an ordered
combinationheads-tails-heads-tailsinthe4coins,(b)theprobabilityofhav-
ing the same ordered combination given that any two coins are known to
have landed heads-up and (c) the probability of having two coins land heads
up given that the sequence heads-tails-heads-tails has occurred.
Solution:
(a) A specific sequence has a probability of P(A) = (1/2)4, since the 4
tosses are independent, and each has a probability 1/2 of landing heads up.
(b) We need to count the number of sequences that have 2 coins landing
heads up (event B). Using the result from Problem 1.1, we see that there
are 6 such sequences, and therefore the probability is P(B)=6/16 by direct
count. The event B includes A, thus the probability of the intersection is
P(A) = 1/16. The probability of a sequence heads-tails-heads-tails given
that two coins land heads up is
P(A∩B) 1 16 1
P(A/B)= = × = .
P(B) 16 6 6
8 1
(c) Use Bayes’ theorem to find
P(A/B)P(B) 1 6 16
P(B/A)= = × × =1.
P(A) 6 16 1
Theknowledgeoftheheads-tails-heads-tailssequenceinfactimpliesthecer-
tainty that 2 coins have landed heads up.
Chapter 2
2.1.Consider the exponential distribution
f(x)=λe−λx
where λ ≥ 0 and x ≥ 0. Show that the distribution is properly normalized,
and calculate the mean, variance and cumulative distribution F(x).
Solution: To show that the exponential distribution is properly normal-
ized, calculate
λ(cid:90) ∞e−λxdx=λ× 1 (cid:16)−e−λx(cid:12)(cid:12)∞(cid:17)=1.
λ 0
0
Use the defition of mean, µ=(cid:82)∞xλe−λxdx, and integrate by parts:
0
µ= −λx1e−λx(cid:12)(cid:12)(cid:12)∞+ λ1 (cid:90) ∞e−λxdx=0− 1e−λx(cid:12)(cid:12)(cid:12)∞ = 1.
λ (cid:12) λ λ (cid:12) λ
0 0 0
The variance is given by σ2 = E[x2]−µ2, in which the moment of the
second order is given by E[x2] = (cid:82)∞x2λe−λxdx. The moment of second
0
order is calculated by integration by parts,
E[x2]=−x2e−λx(cid:12)(cid:12)∞+2(cid:90) +∞xe−λxdx=0+ 2
0 λ2
−∞
and therefore the variance is 1/λ2.
The cumulative distribution is given by
(cid:90) x
F(x)= λe−λtdt=−e−λt(cid:12)(cid:12)x =1−e−λx.
0
0
9
10 2
2.2.Consider the sample mean as a random variable defined by
N
1 (cid:88)
x= x
N i
i=1
where x are identical independent random variables with mean µ and vari-
i
ance σ2. Show that the variance of x is equal to σ2/N.
Solution: The variance of the random variable
N
1 (cid:88)
X = X
N i
i=1
can be calculated using the property that Var(aX) = a2Var(X), the lin-
earity of the expectation operator, and the fact that all variables X are
i
independent:
1 σ2
Var(X)= ·Nσ2 = .
N2 N
2.3.J. J. Thomson’s experiment aimed at the measurement of the ratio be-
tween the mass and charge of the electron is presented on page 23. Using
the datasets for Tube 1 and Tube 2 separately, calculate the mean and vari-
anceoftherandomvariablesW/QandI,andthecovarianceandcorrelation
coefficient between W/Q and I.
Solution: Using the definition of sample mean, variance, covariance and
correlation coefficient, the results are reported in Table2.1. For convenience,
each measurement of W/Q is missing a factor of 1011
Table 2.1 MeasurementsforProblem2.3
N W/Q σ2 σ I¯ σ2 σ σ2 r
W/Q W/Q I I 12
Tube 1 12 13.3 71.5 8.5 312.9 8,715.7 93.4 759.1 0.96
Tube 2 11 2.92 0.67 0.82 174.3 286.2 16.9 13.2 0.95
2.4.Using J. J. Thomson’s experiment (page 23), verify the statement that
‘It will be seen from these tables that the value of m/e is independent of the
nature of the gas’ used in the experiment. You may do so by calculating the
meanandstandarddeviationforthemeasurementsineachgas(air,hydrogen
and carbonic acid) and testing whether the three measurements agree with
each other within their standard deviations.