Cumulative distribution function

Solutions to Quiz 1 and Quiz 2 Fall 2009

Why n people can be arranged in n chairs in a circle in (n-1)! ways?
The total number of arrangements to make three people, say, 1, 2, and 3 to sit in a circle are (3-1)!
= 2! = 2. These two arrangements, are seen by taking one arrangement from each row in the attached
figure. Please note that all the remaining arrangements in the same row are considered equal because
they can be obtained from the preceding  arrangement by going one step in the clockwise direction.

In general, for each of the n! arrangements on a straight line, n clockwise rotations on the circle
are lost in the ratio giving n!/n = 1.(n-1)! = (n-1)! arrangements of n people on n chair in a circle.

Solution to sum of two independent Gammas (X, Y) with parameters (5, λ) and (3, λ), respectively.
To see the full c.d.f. of Gamma (8, λ) see this note.
Page 1,  Page 2.

Practice Problems:

Quiz 2

Solution to Quiz 2.

----------------------------------------------------------------

Exam I:

-----------------------------------------------

Fall 2011 Exam I

Solution

-----------------------------------------------

Fall 2010 Exam I

Solution

------------------------------------------------

Fall 2009 Exam I

Solution

------------------------------------------------

Fall 2007 Exam I Solution

-----------------------------------------------------

Fall 2006 Exam I
.

Solution to fall 2006 Exam I

------------------------------------------------------

Fall 1999 p1

p2

p3

p4

solution to Fall 1999 exam I

-----------------------------------------------------------

Exam I Spring 1996 p1

p2

p3

p4

--------------------------------------------------------------------------------

Exam II:

Solution

---------------------------------------------------------------------------------

Solution

---------------------------------------------------------
Fall 2009 Exam II

Solution

---------------------------------------------------------

Fall 2007 Exam II Solution p1

p2

p3

-------------------------------------------------------

Fall 2006 Exam II.

Solution Fall 2006 p1

-------------------------------------------------------

Math 244-002 p1

p2

p3

p4

p5

Solution to Math 244-002 p1

p2

p3

---------------------------------------------------------------
Fall 1999 p1

p2

p3

p4

Solution to Fall 1999 p1

------------------------------------------------------------

Fall 1996 p1

p2

p3

p4

Solution to Fall 1996 p1

p2

----------------------------------------------------------------
Spring 1993 p1

p2

p3

p4

Solution to Spring 1993 p1

p2

p3
------------------------------------------------

Final Exam:

Fall 2011
Final Exam

Solution

------------------------------------------------

Fall 2010      Final Exam

Solution

-------------------------------------------------

Fall 2009 Final Exam

------------------------------------------------

Fall 2007  Final Exam

Fall 2007 Solution p1

p2

p3

p4

------------------------------------------------

Fall 2006 Final  Exam

---------------------------------------------------------------

Math 244-002

p1

p3

p4

p5
Solution to Math 244-002

p1

p2

p3

p4

p5.