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36201 Spring 2008
Homework 6 – KEY
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View Full Document Stat 201
Spring
2008
Page
1
of
7
Homework
6 – KEY
This document available on:
http://www.stat.cmu.edu/~gordonw/spring2008stat201.html
Due:
Wednesday,
February 27
, at the beginning of lecture.
Homework considered late beyond the first 10 minutes of lecture.
Late Homework Policy:
Late homework may be turnedin for a maximum of half credit, up to 4:00 PM on the same
Wednesday the homework is due. Late homework should be placed in the manila envelope which will
be posted by room 328C, Facilities Management Building (FMS). Late homework should be labeled
with the course as well as your name. Please do not turn in any homework to any mailboxes.
Please include the toppage ‘coversheet’, filledout and stapled to your homework.
Show all work.
This homework covers parts of chapter 6 in the course textbook.
For grading purposes:
Exercise 1
Exercise 2
Exercise 3
Exercise 4
effort
& neatness
TOTAL
19
13
8
40
20
100
NOTE:
To save paper, you are encouraged to print doublesided (‘duplex’ mode).
1.
The U.S. Census Bureau compiles data that describe many features of the population. One of the
questions they ask to women of childbearing age is, “How many children have you had?” Based
on results published for 2004, the probability distribution of the number of children, X, that a
randomlyselected U.S. woman of childbearing age will have had is represented by the following
table:*
Stat 201
Spring
2008
Page
2
of
7
Homework
6 – KEY
This document available on:
http://www.stat.cmu.edu/~gordonw/spring2008stat201.html
X
0
1
2
3
4
5 or more
P(X)
.
45
.
17
.
22
.
11
.
04
.
01
* Available online:
http://www.census.gov/population/socdemo/fertility/cps2004/tab0101.xls
(Probabilities have been rounded to the nearest percent for simplicity.)
1 (a)
How many
variables are represented in the distribution? Is (or are) the variable(s)
discrete
or
continuous
?
Answers:
There is one variable (number of children).
It is discrete (since number of children can only be the ‘separated’ values 0, 1, 2, etc.)
1 (b)
Verify that this is a legitimate probability distribution. (Give brief explanation or show
any calculation you perform.)
Solution:
Each probability is obviously between 0 and 1 (i.e., nonnegative and not greater than 1); and the
probabilities sum to 1: .45 + .17 + .22 + .11 + .04 + .01 = 1
1 (c)
Say in words what “P(X
≠
3)” represents in terms of the actual scenario; then compute
P(X
≠
3). [show work.]
Solution:
“P(X
≠
3)”
represents the probability that a (randomly selected) woman will NOT have 3 children.
(Or equivalently, the proportion of women in the population
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This note was uploaded on 02/01/2009 for the course STAT 36201 taught by Professor Gordon during the Fall '08 term at Carnegie Mellon.
 Fall '08
 gordon

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