Statistics for Physical Sciences: An Introduction (Instructor's Solution Manual) (Solutions)

26 Pages·2012·6.52 MB·English

by  Brian Martin

Solutions to Problems

Problems 1

1.1 If the number on the die is odd, the possible events are: 1H, 1T, 3H, 3T, 5H and 5T. If
the number on the die is even, the possible events are: 2HH, 2HT, 2TH, 2TT, 4HH, 4HT,
4TH, 4TT, 6HH, 6HT, 6TH, 6TT, i.e., 18 elements in the sample space.

1.2 There are 10 possible combinations, given by

SS (i < j)= SS , SS , SS , SS , S S , S S , S S , S S , S S , S S .
i j 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5

1.3 The bins are 0! x <12.5, 12.5 ! x <25.0, etc. with widths 12.5. Frequencies are:

Bin 1 2 3 4 5 6 7 8

Frequency 0 1 5 7 9 11 4 3

and the cumulative distribution is:

Bin <12.5 <25 < 37.5 <50 <62.5 < 75 <87.5 <100
Cumulative frequency 0 1 6 13 22 33 37 40

1.4 First the data are ordered from lowest to highest:

22 27 28 34 34 36 38 39 39 43
45 45 45 53 55 56 56 56 59 59
59 61 63 66 67 67 67 67 68 68
68 73 74 76 76 82 87 90 90 97

The median corresponds to P , for which q = np = 20. Since q is an integer, the median is
0.5
the average of the 20th and 21st values, i.e. 59. For P , we have q= 26.8, so P is the
0.67 0.67
average of the 26th and 27th central bin value, i.e. 67.

1.5 For the unbinned data, x = 58.375 and the standard deviation is s =18.62. For the
binned data x = 58.75 and s =18.82. Shepard’s correction would reduce this to 18.47, but
would not alter the estimate for the mean. The percentage of the unbinned data that falls
within the range x !2s < x < x +2s is 97.5%. If the data followed a normal distribution, the
percentage would be 95.4%.

1.6 The left-hand side of (1.8) for n = 2 is

1 N % 1 N (
µ! = $(x2 "2x#+#2)= $x2 "2#µ+#2.
2 N i i &' N i )*
i=1 i=1

But using the definition of the variance,

$ 1 N '
!2 +d2 = #(x "µ)2 +(µ2 "2*µ+*2)
%&N i ()
i=1
+ 1 N . + 1 N .
= #x2 "2µ2 +µ2 +(µ2 "2*µ+*2 = #x2 "2*µ+*2,
,- N i /0 ,- N i /0
i=1 i=1
1.1
which completes the proof.

1.7 From the Gaussian, f (x)=1 ! 2" and the values of x at half maximum height
max
are given by

# (x!µ)2& 1
exp ! = ,
% (
$ 2"2 ' 2

or, by taking logs, the solutions of the quadratic equation

x2 !2µx+(µ2 !2"2ln2)= 0.

The half-width is therefore the difference in the two roots, i.e.

1/2
#4µ2 !4(µ2 !2"2ln2)% =(8ln2)1/2"= 2.35".
$ &

1.8
1 N 1 ( ) 1 ( )
! = #(x $µ)3 = (x$µ)3 = x3 $3x2µ+3xµ2 $µ3
N"3 i "3 "3
i=1
1 ( ) 1 ( )
= x3 $3x2x +3x x2 $ x3 = x3 $3x2x +2x3 .
"3 "3

1.9 The data give D = 5 and R = 6, from which s = 2.89 and s = 2.80. Also
D R
RD = 25.27, so the covariance is –5.16 and the correlation coefficient is r = !0.64.

Problems 2

2.1 Probability for a current to flow is P =[(1!2)"(3"4)], Using the calculus of
probabilities and noting that the switches act independently, P may be written as

P = P[1!2]+P[3"4]#P[(1!2)!(3"4)]
= P[1!2]+P[3]+P[4]#P[3!4]#P[(1!2)!(3"4)]

= P[1!2]+P[3]+P[4]#P[3!4]#P[(1!2)]{P[3]+P[4]#P[3!4]}
= 2p#2p3 + p4.

The probability for no current to flow is

P = P[1!2]P[3"4]={P[1]+P[2]#P[1"2]}P[3"4].

Using P[I]=1! p, gives P =1!2p+2p3 ! p4 and P+P =1, which checks.

2.2 (a) A!D =(6,9) (b) (B!C)"A=(2,3,4,5,8,9)
(c) (A!B)!C =(4) (d) A!(B"D)=(2,4,8)

2.3 Let F denote the event where the component is faulty and B the event where the
component is part of a faulty batch. We need to calculate

1.2
P[F!B]
P[F|B]= ,
P[B]
where
P[F]= 0.7, P[F]= 0.3, P[B|F]=1 and P[B|F]= 0.15.
Then
P[F!B]= P[F]P[B|F]= 0.7,
and
P[B]= P[B|F]P[F]+P[B|F]P[F]= 0.745.

So P[F|B]= 0.94 and the technician should now be 94% certain that the component is the
problem.

2.4 (a) The probability that team B will win a given game labelled by n is P[B ]= 3 10.
n
The probability of winning both games is therefore

P[B !B ]= P[B ]P[B ]= 9 100,
1 2 1 2

if the results of the two games are independent. (b) Likewise, P[A ]= 5 10. Then,
n

P[A wins at least one game]=1!P[A loses both games]

=1!P[A "A ]=1!1 4 = 3 4.
1 2

2.5 If the books are treated as three blocks, there are 3! possible arrangements. But there
are 5! possible orderings of the physics books within the block, 4! orderings of the maths
books and 3! orderings of the chemistry books. So the number of possible arrangements is

3!5!4!3!= 6!120!24!6=103680.

2.6 (a) The two events are not mutually exclusive, so the probability is

1 1 1 4
P[9!C]= P[9]+P[C]"P[9#C]= + " = .
13 4 52 13

(b) There could be either (i) 3 cards of one suite and one card of a different suite, or (ii) all 4
cards of the same suite. In (i), we multiply the number of combinations of 3 cards having the
same suite by the number of ways the fourth card could be selected from the other three
suites, i.e.

13!
C !39 = !39 =11154.
13 3 10!3!

In (ii), we need the number of combinations of 4 cards of one suite, i.e. C = 715. Thus,
13 4
there are n =11154+715 =11869 combinations that satisfy the criterion. But the total
E
number of possible 4 card-combinations that can be formed from the deck 52 is

52!
n = C = = 270725,
52 4 4!48!

and so the probability is P =11869 270725= 0.044.

1.3
2.7 From the data, 5% of guilty people tested will be judged innocent and 1% of innocent
people tested will be judged guilty. If we denote a positive (guilty) test result by + and a
negative (innocent) test result by –, and guilt and innocence by G and G, then

P[+|G]= 0.95, P[!|G]= 0.05, P[+|G]= 0.01, P[!|G]= 0.99 .

We also know that in the population tested P[G]= 0.03 and P[G]= 0.97. Using these in
Bayes’ theorem gives

P[G]P[+|G]
P[G|+]= = 0.254.
P[G]P[+|G]+P[G]P[+|G]

Thus there is a 25% chance that the suspect has not committed a crime, despite the positive
test.

2.8 In an obvious notation, Bayes’ Theorem is

P[A]P[R|A]
P[A|R]= ,
P[A]P[A|R]+P[B]P[R|B]

with P[R|A]= 2 3 and P[R|B]=1 4. In the absence other information, Bayes’ hypothesis
is P[A]= P[B], so that

P[R|A] 2 3 8
P[A|R]= = = .
P[A|R]+P[R|B] 2 3+1 4 11

2.9 In case (a), the events are not independent and so

P[R !W !B ]= P[R ]P[W |R ]P[B |R !W ]
1 2 3 1 2 1 3 2 2

where W means the second draw has resulted in a white ball etc. Because the number of
2
balls in the bag is reduced by one after each drawing, the right-hand side is

(3 15)!(4 14)!(5 13)= 2 91= 0.022.

In case (b) the events are independent and

P[R !W !B ]= P[R ]P[W ]P[B ].
1 2 3 1 2 3

As the number of balls in the bag remains constant the right-hand side is

(3 15)!(4 15)!(5 15)= 4 225= 0.018.

Problems 3

3.1 The characteristic function is,

a" %
!(t)= &ex($a+it)x"$1dx.
#(")
0

Using the substitution u = x(a!it), and the definition of the gamma function,
1.4
a" % ' it*$"
!(t)= &e$uu"$1du =)1$ ,
#(")(a$it)" ( a+
0
2
it 'it*
=1+" +"(" +1)) , +!
a ( a+

Hence, using (3.13) and (3.15), µ! =" a and µ! ="(" +1) a2 .
1 2

3.2 Consider the random variable x (i =1,2,…,n), where
i

!# 1 if the letter is in correct envelope
x = " .
i
$# 0 otherwise

Then P[x =1]=1 n, and E[x ]=1!P[x =1]+0!P[x = 0]=1 n so that
i i i i

1
E[x + x +!+ x ]= E[x ]+E[x ]+!+E[x ]= n =1.
1 2 n 1 2 n n

Thus, on average, one letter will arrive at its intended destination, independent of n.

3.3 Let x represent a capacitor with magnitude i, so that
i

!# 1 if at least one capacitor of value i is received in 10 deliveries
x = " .
i
$# 0 otherwise

Then, because whether or not particular capacitors are received are independent events, the
expectation value of x is
i

E[x ]= P[that at least one capacitor of size i is received in 10 deliveries]
i
=1!P[no capacitors of size i are received in 10 deliveries]
5
" 9 %
=1!$ ' = 0.4095,
#10&

and hence E[x + x +!+ x ]= E[x ]+E[x ]+!+E[x ]= 4.095.
1 2 10 1 2 10

3.4 The probability distribution is given by the inversion theorem :

1 2! 1 2! 1 2!
p = $ e"ik#cos#d#= $ ei(1"k)#d#+ $ e"i(1+k)#d#.
k 2! 4! 4!
0 0 0

For k = ±1, p =1 2 and for all other k, p = 0, so the probability distribution is
k k

!# 1 2 k = ±1
p = "
k
$# 0 otherwise.

1.5
3.5 (a)
" $y ' "
y
P[x < y]= 3#e!3y&#e!xdx)dy= !3#e!3y$%e!x'( dy= 0.25.
0 %&0 () 0 0
(b)
2 $# ' 3 2
P[x >1,y<2]= 3"e!3y&"e!xdx)dy= "e!3ydy= 0.367.
e
% (
0 1 0

3.6 From the definition of the variance,

!2 = E#{(ax+by)"E[ax+by]}2% = E#{a(x"µ )+b(y"µ )}2%
h $ & $ x y &
= E#a2(x"µ )2 +b2(y"µ )2 +2ab((x"µ )(y"µ )%
$ x y x y &
= a2!2(x)+b2!2(y)+2abcov(x,y).

3.7 The conditional density is fC(x|y)= f(x,y) fM(y), where fM(y) is the marginal
density of y. The latter is given by

1 #x3 x4 x3y&1 c
fM(y)= ! f(x,y)dx = c + " = [7"4y].
% (
3 4 3 12
$ '
0 0
So,
12x2(1+ x! y)
fC(x|y)= .
(7!4y)

3.8
(a) P[x !2,y=1]= P[x =1,y=1]+P[x = 2,y=1]= 19 ,
100
(b) P[x >2,y!2]= P[x = 3,y=1]+P[x = 3,y= 2]= 1 ,
10
(c) P[x+ y= 5]= P[x =1,y= 4]+P[x = 2,y= 3]+P[x = 3,y= 2]= 19 .
100

3.9 From (3.33b),
1
f(u)= ! f (u" x)dx.
y
0

Changing variables by setting v=u! x, gives

u
f(u)= " f (v)dv.
y
u!1

Now 0 <u <2, so we consider two cases: 0 !u <1 and 1!u <2. For the former,

u
f(u)= ! f (v)dv=u, (0"u <1),
y
0
and for the latter,
1
f(u)= " f (v)dv= 2!u, (1#u <2).
y
u!1

1.6
Plotting these shows that the resulting distribution is triangular with a peak at u =1.

Problems 4

4.1 Without medical intervention patients either do, or do not, recover, so the binomial
distribution is appropriate with p = 0.5. If n is the number of patients who recover without
using the drug, then

! 20 $
P[n=14]= '0.514 '0.56 = 0.0369,
# &
" 14 %

i.e. 3.7%. This is to be compared with 70% who recover if they use the drug. So, on the basis
of this very small sample, the drug is effective.

4.2 Using the binomial distribution with r as the number of hits, we need to find the value
of n such that P[r !2;n,0.95]!0.999. We have,

2 ! 2 $
n = 2: P = ' (0.95)i(0.05)2(i = 0.9025,
# &
" i %
i=2
3 ! 3 $
n = 3: P = ' (0.95)i(0.05)3(i = 0.9927,
# &
" i %
i=2
4 ! 4 $
n = 4: P = ' (0.95)i(0.05)4(i = 0.9995.
# &
" i %
i=2

So the minimum number of layers is four.

4.3 Because the sum of the probabilities must add to unity, we have

P[h]= 0.48, P[t]= 0.49, P[e]= 0.03.

Then from the multinomial distribution, the required probability is

10!
P = (0.48)4(0.49)4(0.03)2 = 0.0087.
4!4!2!

4.4 Using the multinomial distribution, the general term in the expansion is

! 6! $ ! 6! $
(2x2)r1('3xy2)r2(z3)r3 = 2r1('3)r2 x2r1+r2y2r2z3r3 ,
# & # &
"r!r !r !% "r!r !r !%
1 2 3 1 2 3

where (r +r +r )= 6. We need 2r +r = 6, 2r = 4 and 3r = 6, i.e. r = r = r = 2, and
1 2 3 1 2 2 3 1 2 3
so the coefficient is
6!
22(!3)2 = 3240.
2!2!2!

4.5 As an incident particle either does, or does not, produce a new particle, the binomial
distribution may be used. If r is the number of emitted particles per second, then

1.7
4
P[r !5]=1"#P[r =i],
i=0
where
! 100 $
P[r]= (0.05)r(0.95)100'r.
# &
" r %
This gives
P[0]= 0.0059, !P[1]= 0.0312, P[2]= 0.0812,

P[3]= 0.1396, P[4]= 0.1781,
and so
P[r !5]= 0.564.

Because p is small and the number of potential events is large, we can use the Poisson
approximation to the binomial with != np= 5. Then,

4 #r
P[r !5]=1"$ e"# = 0.559.
r!
r=0

4.6 If we assume that many cars use the junction, and the probability of any one of them
having an accident is very small, we can use the Poisson distribution to find

3
P[n > 3]=1!"P[n = r],
r=0

where n is the number of accidents next month. This gives 0.735.

4.7 A detector will either fail during a run or not, with the former having a probability
p= 0.01, so strictly this situation is described by a binomial distribution. But since p is small
and the number of detectors is large, we can use the Poisson approximation with
!= np = 0.8. We require the probability that at least 3 detectors fail, i.e.

P[k ! 3]=1"P[k = 0]"P[k =1]"P[k = 2]= 0.047.

Thus about 5% of runs will be spoiled because of detector failure.

4.8 In standard measure, R= 48.0!z = "2.0 and R= 451.5!z =1.5. So the
probability that a resistor is defective is

P = P[z <!2.0]+P[x >1.5]=[1! N(2.0)]+[1! N(1.5)]= 0.0896.

4.9 P[W >6!105]"P[50V2 >6!105]"P[V >109.5]. Converting to standard
measure, z =(W !µ) "=1.34 and so

P[W >6!105]"P[z >1.34]=1#P[z <1.34]$0.09.

Problems 5

5.1 By the central limit theorem,

E[x ]=µ= I +1 2 and var(x )=!2 n=112n.
n n
1.8
So, an unbiased estimator for I is x !1 2 and the mean squared error on the biased estimator
n
x is, from (5.9),
n

E[(x !µ)2]= var(x )+b2 =112n+1 4.
n n

5.2 Squaring both sides of the identity

ˆ ˆ ˆ ˆ ˆ ˆ
(! "!)=(! "E[! ])+(E[! ]"!)=(! "E[! ])+b,
n n n n n n
gives
(!ˆ "!)2 =(!ˆ "E[!ˆ ])2 +2b(!ˆ "E[!ˆ ])+b2,
n n n n n

and then taking expectation values,

E[(!ˆ "!)2]= E[(!ˆ "E[!ˆ ])2]+2bE[(!ˆ "E[!ˆ ])]+b2.
n n n n n

The first term on the right-hand side is var("!) and the second term is zero, so
n

E[(!ˆ "!)2]= var(!#)+b2.
n n

5.3 Let x ! x " x . Then from (5.31), µ =µ !µ =10, and from (5.32),
1 2 x 1 2

!2 =!2 n +!2 n = 4.
x 1 1 2 2

Because the population distribution are both normal, the central limit theorem may be used
even for these small sample sizes. Using the normal distribution with standardised variable

z =(x!µ) "=(x!10) 2,
gives
P[x < 7.5]= P[z <!1.25]=1! N(1.25)= 0.1056.

5.4 Let f(x ,x ,…,x ) be the joint density function of x ,x ,…,x . Then from (3.26a),
1 2 n 1 2 n

" " n " " & n ) n
µ = #…# l f(x ,x ,…,x )$dx = #…# %a x $dx
l 1 2 n i ( j j+ i
' *
!" !" i=1 !" !" j=1 i=1
n " " n n
= %a #…# x f(x ,x ,…,x )$dx = %a µ.
jj j 1 2 n i jj j
j=1 !" !" i=1 j=1

Also, using equations (3.26b) and (3.28),

1.9
2
# # n # # ' n * n
!2 = $…$ (l"µ )2f(x ,x ,…,x )%dx = $…$ )&a (x "µ ), f(x ,x ,…,x )%dx
l l 1 2 n i jj j j 1 2 n i
"# "# i=1 "# "#(j=1 + i=1
n # # n
= &a2 $…$ (x "µ )2f(x ,x ,…,x )%dx
j j j 1 2 n i
j=1 "# "# i=1
n # # n
+2&a a $…$ (x "µ )(x "µ )f(x ,x ,…,x )%dx
k j k k j j 1 2 n i
k<j "# "# i=1
n
= &a2!2 +2&a a! .
i i k j kj
i=1 k<j

which completes the proof.

5.5 If P is the probability of sustaining damage after n exposures, then
n

# n &
P = P[W !0] where W = "r ) R.
n n n $% i'(
i=1

By the central limit theorem, the sum is normally distributed with mean 4n and variance 2n,
so the mean and variance of W are (4n!100) and (2n!10), respectively. Thus the
n
variable

W !(4n!100)
z = n
n (2n!10)1/2
is distributed as N(0,1) and
# "(4n"100)&
P = P z ! .
n $% n (2n"10)1/2 '(

With P = 0.03, z =1.88, so that
n n

(100!4n)
"1.88.
(2n!10)1/2

The smallest number for which this is satisfied is n = 23, so any more than 22 exposures will
result in permanent damage with a probability of at least 3%.

5.6 We need P[x ! 49], where x is the mean of the sample A. Using the central limit
A A
theorem and assuming a normal approximation for the sampling distribution, with a standard
deviation ! n = 0.7, the standardised variable z =(x !50) 0.7, gives
A

P[x ! 49]= P[z !"1.43]=1" N(1.430)= 0.0764.
A

This is rather a low value, so it would be worth taking other samples. If the result was
confirmed, then it is likely that the product from plant A is not satisfactory.

5.7 We seek the value of n such that

P$ R ! R "#&= P$R ( R!#&+P$R " R+#&(),
% n ' % n ' % n '
1.10