Chapter 3 Stats
Chapter 3 Fundamental tools for statistics
Descriptive Statistics: summarize or describe relevant characteristics of data
Inferential Statistics: make inferences or generalizations about a population
Measures of Center
o Measure of Center:
From these expressions, its hard to tell which substitution creates the most reliable circuit. First, we
observe that P[W I I ] > P[W I ] if and only if
1
q2
q
q
+ (1 q)3 > 1 (5 4q + q 2 ).
2
2
2
(11)
Some algebra will show that P[W I I ] > P[W I ] if and
By the law of total probability,
P [K 2 ] = P [K 2 |C11 ] P [C11 ] + P [K 2 |C12 ] P [C12 ]
+ P [K 2 |C21 ] P [C21 ] + P [K 2 |C22 ] P [C22 ]
11 11 1 5
7
1 1
+
+
+
= .
=
2 12 3 4 2 4 3 12
18
(12)
(13)
We observe that K 1 and K 2 are not independent since
The probability that they win 10 titles in 11 years is
11
(.32)10 (.68) = 0.00082.
10
P 10 titles in 11 years =
(2)
The probability of each of these events is less than 1 in 1000! Given that these events took place in
the relatively short fty year history
Problem 1.9.5 Solution
(a) There are 3 group 1 kickers and 6 group 2 kickers. Using G i to denote that a group i kicker
was chosen, we have
P [G 2 ] = 2/3.
(1)
P [G 1 ] = 1/3
In addition, the problem statement tells us that
P [K |G 1 ] = 1/2
P [K |G 2 ] =
Finally, with the swingman on the bench, we choose 1 out of 3 centers, 2 out of 4 forward,
and 2 out of four guards. This implies
N3 =
3
1
4
2
4
= 108,
2
(3)
and the number of total lineups is N1 + N2 + N3 = 252.
Problem 1.8.7 Solution
What our design mus
Problem 1.8.5 Solution
When the DH can be chosen among all the players, including the pitchers, there are two cases:
The DH is a eld player. In this case, the number of possible lineups, N F , is given in Problem 1.8.4. In this case, the designated hitte
(c) The probability that the two cards are of the same type but different suit is the number of
outcomes that are of the same type but different suit divided by the total number of outcomes
involved in picking two cards at random from a deck of 52 cards.
Problem 1.7.10 Solution
The experiment ends as soon as a sh is caught. The tree resembles
p C1
p C2
p C3
c
c
c
C3
C1
C2
1 p
1 p
1 p
.
From the tree, P[C1 ] = p and P[C2 ] = (1 p) p. Finally, a sh is caught on the nth cast if no sh
were caught on the pre
For n = 100 packets, the packet success probability is inconclusive. Experimentation will show
that C=97, C=98, C=99 and C=100 correct packets are typica values that might be observed. By
increasing n, more consistent results are obtained. For example, re
(1-q)
3
2
1-q
1-q
The probability P[W ] that the two devices in parallel work is 1 minus the probability that neither
works:
(3)
P W = 1 q(1 (1 q)3 ).
Finally, for the device to work, both composite device in series must work. Thus, the probability
the de
The M ATLAB built-in function ndgrid facilitates plotting a function g(x, y) as a surface over the
x, y plane. The x, y plane is represented by a grid of all pairs of points x(i), y( j). When x has
n elements, and y has m elements, ndgrid(x,y) creates a g
Problem 2.2.6 Solution
The probability that a caller fails to get through in three tries is (1 p)3 . To be sure that at least
95% of all callers get through, we need (1 p)3 0.05. This implies p = 0.6316.
Problem 2.2.7 Solution
In Problem 2.2.6, each calle
Problem 2.3.7 Solution
Since an average of T /5 buses arrive in an interval of T minutes, buses arrive at the bus stop at a
rate of 1/5 buses per minute.
(a) From the denition of the Poisson PMF, the PMF of B, the number of buses in T minutes, is
PB (b) =
Problem 2.3.4 Solution
(a) Let X be the number of times the frisbee is thrown until the dog catches it and runs away.
Each throw of the frisbee can be viewed as a Bernoulli trial in which a success occurs if the
dog catches the frisbee an runs away. Thus,
(c) Let B denote the event that a busy signal is given after six failed setup attempts. The probability of six consecutive failures is P[B] = (1 p)6 .
(d) To be sure that P[B] 0.02, we need p 1 (0.02)1/6 = 0.479.
Problem 2.3.1 Solution
(a) If it is indeed
(c) The probability that V is even is
P [V is even] = PV (2) + PV (4) =
42
2
22
+
=
30 30
3
(3)
(d) The probability that V > 2 is
P [V > 2] = PV (3) + PV (4) =
42
5
32
+
=
30 30
6
(4)
Problem 2.2.4 Solution
(a) We choose c so that the PMF sums to one.
7c
Problem Solutions Chapter 2
Problem 2.2.1 Solution
(a) We wish to nd the value of c that makes the PMF sum up to one.
PN (n) =
Therefore,
2
n=0
c(1/2)n n = 0, 1, 2
0
otherwise
(1)
PN (n) = c + c/2 + c/4 = 1, implying c = 4/7.
(b) The probability that N 1
> ultrareliable6(10000,0.2)
ans =
8738
8762
8806
> > ultrareliable6(10000,0.2)
ans =
8771
8795
8806
>
9135
8800
8796
9178
8886
8875
In both cases, it is clear that replacing component 4 maximizes the device reliability. The somewhat
complicated solution o
Problem 1.10.4 Solution
From the statement of Problem 1.10.1, the device components are congured in the following way:
W1
W2
W3
W5
W4
W6
By symmetry, note that the reliability of the system is the same whether we replace component 1,
component 2, or compo
Similarly,
P [T3 ] = P [T1 H2 T3 H4 ] + P [T1 H2 T3 T4 ] + P [T1 T2 T3 ]
= 1/8 + 1/16 + 1/16 = 1/4.
(3)
(4)
(c) The event that Dagwood must diet is
D = (T1 H2 T3 T4 ) (T1 T2 H3 T4 ) (T1 T2 T3 ).
(5)
The probability that Dagwood must diet is
P [D] = P [T1
2. If we need to check whether the rst resistance exceeds the second resistance, an event space
is
B2 = cfw_R1 R2 .
(2)
B1 = cfw_R1 > R2
3. If we need to check whether each resistance doesnt fall below a minimum value (in this case
50 ohms for R1 and 10
Once again, using Theorem 1.4, we have
P[A B] = P[AB] + P[AB c ] + P[Ac B]
(12)
Substituting the results of Equation (10) into Equation (12) yields
P [A B] = P [AB] + P [A] P [AB] + P [B] P [AB] ,
(13)
which completes the proof. Note that this claim requi
Problem 1.4.6 Solution
(a) For convenience, let pi = P[F Hi ] and qi = P[V Hi ]. Using this shorthand, the six unknowns
p0 , p1 , p2 , q0 , q1 , q2 ll the table as
F
V
H0 H1 H2
p0 p1 p2 .
q0 q1 q2
(1)
However, we are given a number of facts:
p0 + q0 = 1/3
For the mutually exclusive events B1 , . . . , Bm , let Ai = Bi for i = 1, . . . , m and let Ai = for
i > m. In that case, by Axiom 3,
P [B1 B2 Bm ] = P [A1 A2 ]
m1
P [Ai ] +
=
P [Ai ]
(2)
P [Ai ] .
(3)
i=m
i=1
m1
=
(1)
P [Bi ] +
i=m
i=1
Now, we use Axiom
Problem 1.4.7 Solution
It is tempting to use the following proof:
Since S and are mutually exclusive, and since S = S ,
1 = P [S ] = P [S] + P [] .
(1)
Since P[S] = 1, we must have P[] = 0.
The above proof used the property that for mutually exclusive set
Problem 1.4.3 Solution
The rst generation consists of two plants each with genotype yg or gy. They are crossed to produce
the following second generation genotypes, S = cfw_yy, yg, gy, gg. Each genotype is just as likely
as any other so the probability of
Problem 1.2.3 Solution
The sample space is
S = cfw_A, . . . , K , A, . . . , K , A, . . . , K , A, . . . , K .
(1)
The event H is the set
H = cfw_A, . . . , K .
(2)
Problem 1.2.4 Solution
The sample space is
1/1 . . . 1/31, 2/1 . . . 2/29, 3/1 . . . 3/
Problem 1.2.1 Solution
(a) An outcome species whether the fax is high (h), medium (m), or low (l) speed, and whether
the fax has two (t) pages or four ( f ) pages. The sample space is
S = cfw_ht, h f, mt, m f, lt, l f .
(1)
(b) The event that the fax is