CHAPTER 1
1.1 Let ru(k ) = E [u(n)u*(n k )] r y(k ) = E [ y(n) y*(n k )] We are given that y(n) = u(n + a) u(n a) Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get r y(k ) = E [(u(n + a) u(n a)(u*(n + a k ) u*(n a k )] = 2ru(k ) ru(2a +
Key Features of Budget 2016-2017
INTRODUCTION
Growth of Economy accelerated to 7.6% in 2015-16.
India hailed as a bright spot amidst a slowing global economy by IMF.
Robust growth achieved despite very unfavourable global conditions
and two consecutive ye
Adding the values of the seventh column, we obtain Step 6. Obtain the
sample variance by dividing by n 1. Thus, Step 7. Obtain the standard
deviation by taking the positive square root of the variance. Self-Review
Test 1. The value of the middle term in a
Find the (approximate) value of the 60th percentile. Give a brief
interpretation of this value. c. Calculate the percentile rank of 12. Give a
brief interpretation of this value. 23. Make a box-and-whisker plot for
the data on the number of times passenge
employees. JWCL216_ch03_079-136.qxd 12/10/09 9:41 AM Page 121
3.119 Refer to the data of Exercise 3.109 on the current annual incomes
(in thousands of dollars) of the 10 members of the class of 2000 of the
Metro Business College who were voted most likely
the probability of as Figure 4.13 shows a Venn diagram for this example.
P1A2 1 P1A2 1 .70 .30 P1A2 P1A2 35005000 .70 and P1A2 15005000 .
30 A A A A the selected adult is either against such laws or has no
opinion A the selected adult is in favor of stric
spreadsheet within the current Excel workbook, or a new Excel
workbook. 4. Click Summary Statistics. Click OK (see Screen 3.11 for an
example of the output). 5. The Analysis ToolPak does not calculate the
first and third quartiles. To do this, go to an em
had two cards each, 77 had three cards each, 43 had four cards each,
and 410 had five or more cards each. Write the frequency distribution
table for the number of credit cards an adult possesses. Calculate the
relative frequencies for all categories. Supp
an experiment may have the same probability of occurrence. Such
outcomes are called equally likely outcomes. The classical probability
rule is applied to compute the probabilities of events for an experiment
for which all outcomes are equally likely. 144
summary, which is shown in Screen 3.2. 2. Constructing a box-andwhisker plot is similar to constructing a histogram. First enter your data
into a list such as L1, then select STAT PLOT and go into one of the three
plots. Make sure the plot is turned on. F
The probability of the intersection of two events is called their joint
probability. It is written as P1A and B2 The probability of the
intersection of two events is obtained by multiplying the marginal
probability of one event by the conditional probabil
test contains two multiple-choice questions. If a student makes a
random guess to answer each question, how many outcomes are
possible? Depict all these outcomes in a Venn diagram. Also draw a tree
diagram for this experiment. (Hint: Consider two outcomes
probability of A given that B has already occurred. P1A 0 B2
JWCL216_ch04_137-190.qxd 11/17/09 4:51 PM Page 151 EXAMPLE 4
15 Compute the conditional probability P(in favor | male) for the data
on 100 employees given in Table 4.4. Solution The probability
11/17/09 4:51 PM Page 152 Hence, the required probability is The tree
diagram in Figure 4.7 illustrates this example. P1female 0 in favor2
Number of females who are in favor Total number of employees who
are in favor 4 19 .2105 Figure 4.7 Tree diagram. Ma
considered at a time, the employee selected can be a male, a female, in
favor, or against. The probability of each of these four characteristics or
events is called marginal JWCL216_ch04_137-190.qxd 11/17/09 4:51
PM Page 150 probability or simple probabil
CHAPTER 5
5.1 From Fig. 5.2 of the text we see that the LMS algorithm requires 2M+1 complex multiplications and 2M complex additions per iteration, where M is the number of tap weights used in the adaptive transversal lter. Therefore, the computational co
CHAPTER 4
4.1 (a) For convergence of the steepest-descent algorithm: 2 0 < < - max where max is the largest eigenvalue of the correlation matrix R. We are given 1 0.5 0.5 1
R=
The two eigenvalues of R are 1 = 0.5 2 = 1.5 Hence max = 1.5 . The step-size pa
CHAPTER 2
2.1 (a) Let wk = x + jy p(-k) = a + jb We may then write f = wk p*(-k) = (x + jy)(a - jb) = (ax + by) + j(ay - bx) Let f = u + jv with u = ax + by v = ay - bx Hence, u - = a x v - = a y u - = b y v - = b x
From these results we immediately see t
CHAPTER 3
3.1 (a) Let aM denote the tap-weight vector of the forward prediction-error lter. With a tapinput vector uM+1(n), the forward prediction error at the lter output equals f M ( n ) = a M u M +1 ( n ) The mean-square value of fM(n) equals E [ f M (
Compare the values of the various statistics for different sectors. Create
a stacked dotplot of the highest prices for various sectors with each
sectors data as one set of data. Explain how the results of your
comparisons can be seen in the dotplot. TA3.8
to grams)? Does the same thing happen if you convert from a smaller
unit (e.g., pounds) to a larger unit (e.g., stone)? Supplementary
Exercises 125 JWCL216_ch03_079-136.qxd 12/10/09 9:44 AM Page 125
d. Figure 3.15 gives a stacked dotplot of these weights
Solution Because the selection is to be made randomly, each of the 500
women has the same probability of being selected. Consequently this
experiment has a total of 500 equally likely outcomes. One hundred
twenty of these 500 outcomes are included in the
determine the premium for this insurance? 3.137 A local golf club has
mens and womens summer leagues. The following data give the scores
for a round of 18 holes of golf for 17 men and 15 women randomly
selected from their respective leagues. 124 Chapter 3
which are compound events. a. At least one part is good. b. Exactly one
part is defective. c. The first part is good and the second is defective. d.
At most one part is good. 4.14 Refer to Exercise 4.8. List all the
outcomes included in each of the follow
outcomes for an experiment that are not in A. A Figure 4.11 Venn
diagram of two complementary events. A S A Because two
complementary events, taken together, include all the outcomes for an
experiment and because the sum of the probabilities of all outcom
once put it, is enough to make a baseball player toss out his lucky bat or
start seriously searching for flaws in his hitting technique. But the
culprit is usually just simple mathematics. Statistician Harry Roberts of
the University of Chicagos Graduate
because whether or not one fire detector goes off during a fire has no
effect on the second fire detector. We define the following two events:
B the second fire detector fails to go off during a fire A the first fire
detector fails to go off during a fire
of the squared deviations by n 1. Thus Step 6. Obtain the sample
standard deviation by taking the positive square root of the variance.
Hence, s 16737.80 82.0841 $82.08 billion s 2 1x x 2 2 n 1 26,951.20 5
1 6737.80 1x x2 2 26,951.20 1x x 2 2 ; 1x x2 2 x
lower death rate in both age groups have the higher overall death rate?
(This phenomenon is known as Simpsons paradox.) 2.4324 $2.35 12.22
2.34 2.41 Supplementary Exercises 123 JWCL216_ch03_079-136.qxd
12/10/09 9:42 AM Page 123 3.132 In a study of distanc
for parking spaces on the first day of classes in the Fall semester of
2009. Time Number of Students 0 to less than 4 1 4 to less than 8 7 8 to
less than 12 15 12 to less than 16 18 16 to less than 20 6 20 to less than
24 3 Find the mean, variance, and st
obtain before making a decision about the suburb where you should
buy a house? By the time you get to college, you must have heard it
over and over again: A picture is worth a thousand words. Now we
have pictures and numbers discussed in Chapters 2 and 3,
data set such that about a. 75% of the values are smaller and about 25%
are larger than this value b. 25% of the values are smaller and about
75% are larger than this value 15. The following data give the numbers
of times 10 persons used their credit card
have applied for a security guard position with a company. Of them, 7
have previous experience in this area and 25 do not. Suppose one
applicant is selected at random. Consider the following two events: This
applicant has previous experience, and this app