BE200
1
8
Problem Set 9
Problem 1
Given:
Population mean X = 4.38
Population standard deviation X = 0.06
Sample size n = 10
a) The sample mean MX is distributed as Gaussian with mean equal to MX and standard
deviation equal to X .
n
b) The expected value

Review 3
12/02/2013
Exam 3
Material
Markov Processes
1-sample t tests and z tests
Office hours:
Monday, 2:30-3:30
Tuesday, 4:30-5:30
Both in 36 Cummington St, Room 304
Problem 1
A certain protein has three configurations, which we
can denote as C1,

BE200
Solution 5
Problem 1: From problem set 4 problem 1,we have distribution as below.
xi
fx (xi )
2
x=0
E [X ] = =
0
0.25
1
0.5
2
0.25
xi fx (xi ) = (0)(0.25) + (1)(0.5) + (2)(0.25) = 1
V ar[X ] = 2 = E [X 2 ] 2
x
E [X 2 ] =
2
x=0
x2 fx (xi ) = (0)(0.25

BE200
Solution 4
Problem 1:
a)
Figure 1: Probability Function
b)
Figure 2: Cumulative Distribution Function
Problem 2:
a) We know that
3
0
fX (x)dx =
1
c= 9
2
fX (x) = x
9
3
0
+
fX (x)dx = 1, then
3
cx2 dx = c x |3 = c9 = 1
30
1
BE200
Solution 4
b)P (1 <

BE200
Problem set 7
8
Problem 1: a:
V = [1, 1 , 1 ] is not a state vector because
vi > 1 and vi < 0.
2
2
1
vi > 1.
V = [ 2 , 1 , 1 ] is not a state vector because
22
b:
V = [1, 1, 1]
333
V = [1, 1, 1]
632
c:
Dene pi,j is the element at the row i and colum

BE200, Oct 2, 2006
1
Review Session 1
Problem 1
While this problem can be solved directly using conditional probability, it is useful to note that,
this is a Binomial experiment. If we consider each toss as a trial, and a head as a success, then,
The numb

BE200 Review 2
10/28/2013
Office Hours this week
Tuesday:
4:30-6:30pm
36 Cummington St, Room 304
Things to keep in mind
For any probability (density) function, the sum (integral) of it s
hould be 1.
For any cumulative distribution function, F()=1.
Wh