Download ECE316 Tutorial for the week of June 29

ECE316 Tutorial for the week of June 29-July 3
Problem 5.7 – Page 225
The density function of X is given by
0
1
0
If E[X] = 3/5, find a and b.
Solution:
Since we have 2 unknowns, we need to have 2 equations.
1
1st equation :
2nd equation :
,
Solving the 2 equations, we will get ...
Problem 5.14 – Page 225
Let X be a uniform (0,1) random variable. Compute E[Xn] by using proposition 2.1, and then
check the result by using the definition of expectation.
Solution:
Ö Using the definition of expectation :
Let
, for 0
1
0
1 then the CDF will be
/
Now we need to find the PDF …
Then
/
/
Ö Using proposition 2.1 :
Problem 5.15 – Page 225
10 and
If X is a normal random variable with the parameters
36, compute
5
(a)
(b)
(c)
(d)
(e)
4
16
8
20
16
Solution:
, where Z have a zero mean and
Define a normalized normal random variable
unity variance.
Now, we can use Table 5.1 to find the probabilities.
5
(a)
4
(b)
16
1
Then,
.
∞
1
0.8413
1
1
4
16
(c)
8
(d)
20
(e)
16
1
1
0.8413
1
1
0.8413
1
0.7967
1
1
1 0.8413
0.6826
1
1
0.6293
0.3707
0.9525
1
1
1
1
0.8413
0.1587
Problem 5.16 – Page 225
The annual rainfall (in inches) in a certain region is normally distributed with
40, and
4
What is the probability that, starting with this year, it will take over 10 years before a year occurs
having a rainfall of over 50 inches? What assumptions are you making?
Solution:
50
2.5
0.9938
Assuming that the rainfalls in different years are independent, then the Probability that it will
take 10 years before having rainfall over 50 inches is = 0.9938
Problem 5.17 – Page 225
A man aiming at a target receives 10 points if his shot is within 1 inch of the target, 5 points if
between 1 and 3 inches of the target, 3 points if between 3 and 5 inches if the target. Find the
expected number of points scored if the distance from the shot to the target is uniformly
distributed between 0 and 10.
Solution:
Let X be a random variable representing the distance between the shot and the target.
1
0.1
0.1
1
3
0.1
0.2
5
3
0.1
0.2
0.1
10
0.2
5
0.2
3
2.6
Problem 5.18 – Page 225
Suppose that X is a normal random variable with mean 5. If
what is the Var(X)?
9
Solution:
1
9
0.2
0.8
Using table, we can see that
0.84, from which we can get
4.76
0.2, approximately