EP INTRODUCTION TO PROBABILITY+STAT.
14th Edition
ISBN: 2810019974203
Author: Mendenhall
Publisher: CENGAGE L
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Chapter 15.3, Problem 15.17E
(a)
To determine
To find: Whether there is sufficient evidence to conclude that one of the assessors is consistently more conservative than the other.
(b)
To determine
To compare: The answer the answer of part (a) with the answer of exercise 10.46 and conclude whether the test results agree and consistent or not.
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II Consider the following data matrix X:
X1
X2
0.5
0.4
0.2 0.5
0.5
0.5
10.3 10
10.1 10.4
10.1
10.5
What will the resulting clusters be when using the k-Means method with k = 2. In your own words, explain why this result is
indeed expected, i.e. why this clustering minimises the ESS map.
why the answer is 3 and 10?
PS
9 Two films are shown on screen A and screen B at a cinema each evening. The numbers
of people viewing the films on 12 consecutive evenings are shown in the back-to-back
stem-and-leaf diagram.
Screen A (12) Screen B (12)
8
037
34
7 6 4 0 534
74 1645678
92 71689
Key: 116|4 represents 61 viewers for A and 64 viewers for B
A second stem-and-leaf diagram (with rows of the same width as the previous diagram)
is drawn showing the total number of people viewing films at the cinema on each of
these 12 evenings. Find the least and greatest possible number of rows that this second
diagram could have.
TIP
On the evening when 30 people viewed films on screen A, there could have been as few
as 37 or as many as 79 people viewing films on screen B.
Chapter 15 Solutions
EP INTRODUCTION TO PROBABILITY+STAT.
Ch. 15.2 - Prob. 15.1ECh. 15.2 - Prob. 15.2ECh. 15.2 - Prob. 15.3ECh. 15.2 - Prob. 15.4ECh. 15.2 - Prob. 15.5ECh. 15.2 - Prob. 15.6ECh. 15.2 - Prob. 15.7ECh. 15.2 - Prob. 15.8ECh. 15.2 - Prob. 15.10ECh. 15.2 - Prob. 15.12E
Ch. 15.3 - Prob. 15.13ECh. 15.3 - Prob. 15.14ECh. 15.3 - Prob. 15.15ECh. 15.3 - Prob. 15.16ECh. 15.3 - Prob. 15.17ECh. 15.3 - Prob. 15.18ECh. 15.3 - Prob. 15.19ECh. 15.3 - Prob. 15.20ECh. 15.5 - Prob. 15.21ECh. 15.5 - Prob. 15.22ECh. 15.5 - Prob. 15.23ECh. 15.5 - Prob. 15.24ECh. 15.5 - Prob. 15.25ECh. 15.5 - Prob. 15.26ECh. 15.5 - Prob. 15.28ECh. 15.5 - Prob. 15.30ECh. 15.6 - Prob. 15.32ECh. 15.6 - Prob. 15.33ECh. 15.6 - Prob. 15.34ECh. 15.6 - Prob. 15.35ECh. 15.6 - Prob. 15.37ECh. 15.7 - Prob. 15.38ECh. 15.7 - Prob. 15.39ECh. 15.7 - Prob. 15.40ECh. 15.7 - Prob. 15.41ECh. 15.7 - Prob. 15.42ECh. 15.8 - Prob. 15.43ECh. 15.8 - Prob. 15.44ECh. 15.8 - Prob. 15.45ECh. 15.8 - Prob. 15.46ECh. 15.8 - Prob. 15.50ECh. 15.8 - Prob. 15.51ECh. 15.8 - Prob. 15.53ECh. 15.8 - Prob. 15.54ECh. 15 - Prob. 15.55SECh. 15 - Prob. 15.56SECh. 15 - Identical Twins to compare two junior high...Ch. 15 - Identical Twins II Refer to Exercise 15.57. What...Ch. 15 - Prob. 15.59SECh. 15 - Meat Tenderizers An experiment was conducted to...Ch. 15 - Prob. 15.62SECh. 15 - Prob. 15.63SECh. 15 - Prob. 15.65SECh. 15 - Prob. 15.66SECh. 15 - Yield of Wheat Exercise 11.68 presented ananalysis...Ch. 15 - Prob. 15.68SECh. 15 - Prob. 15.69SECh. 15 - Prob. 15.70SECh. 15 - Prob. 15.71SECh. 15 - Prob. 15.72SECh. 15 - Prob. 15.73SECh. 15 - Prob. 15.74SECh. 15 - Prob. 15.75SECh. 15 - Prob. 15.77SECh. 15 - Prob. 15.78SECh. 15 - Prob. 15.79SECh. 15 - Prob. 1CSCh. 15 - Prob. 2CSCh. 15 - Prob. 3CS
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