Unformatted text preview: ECE 316 for the week of June 2226, 2009
Problem 1 – Page 224:
Let X be a random variable with probability density function
1
1
otherwise 1
0
a) What is the value of c?
b) What is the cumulative distribution function of X? Solution:
1 a)
1 1 1 1 b) 1 1 1
0 1
, 1 1 1
1 Problem 2 – Page 224:
A system consisting of one original unit plus a spare can function for a random amount of time X. If the
density of X is given (in units of months) by
/ 0 0
0 what is the probability that the system functions for at least 5 months? Solution:
1
/ 1 By using and letting
, 0 / 2
/ 2
4 / 2
/ / and 1 1
5 The probability that the system functions for at least 5 months
/ 5
/ 2 4 / 10 / 4 / / Problem 3 – Page 224:
Consider the function
2
0 5
2
otherwise
0 Could f be a probability density function? If so, determine C. Repeat if f(x) were given by
2
0 5
2
otherwise
0 Solution:
In order f to be a probability density function, it should be positive for all x values.
2 2
Roots of 2 is 0, √2 f (x) + ve x ve 5/2
0
√2
since f(x) has a negative values from √2 to 5/2, therefore f(x) could not be a probability density function.
2 And for 2
Roots of 2 f (x) is 0, 2 + ve x ve 0
5/2
2
since f(x) has a negative values from 2 to 5/2, therefore f(x) could not be a probability density function.
Therefore, if the second root is greater than 5/2, the function can be a probability density function. Problem 4 – Page 224:
The probability density function of X, the lifetime of a certain type of electronic device (measured in
hours), is given by
10 10 0 10 a) Find
20
b) What is the cumulative distribution function of X?
c) What is the probability that, of 6 such types of devices, at least 3 will function for at least 15
hours? What assumptions are you making? Solution:
a) 20
1 b)
0
1 , 10
10 c) Probability that one device will function at least 15 hours =
15 10 15 Probability that, of 6 such types of devices, at least 3 will function for at least 15 hours is
∑ Problem 5 – Page 224:
A filling station is supplied with gasoline once a week. If its weekly volume of sales in thousands of
gallons is a random variable with probability density function
51
0 0
1
otherwise what must the capacity of the tank be so that the probability of the supply’s being exhausted in a given
week is 0.01?
Solution:
Let c be the capacity of the tank.
0.01 51
1 0.01, 1 0.01 / 0.6 Problem 6 – Page 224:
Compute if X has a density function given by
/ a) 0 0
1
0
0 b)
c) otherwise
1
1
otherwise
5
5 Solution:
/ a)
By using
2 and letting
, 2 / and / / 2 / 4 / By applying again and letting , 2
/ 2
2
b) / / 2 / and 4 / / 2 / 4 1
0 Note:
1,1 .
c) is an odd function, i.e. symmetric around the origin, which will lead to zero area along 5ln  ∞ ...
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This note was uploaded on 12/22/2011 for the course CS 101 taught by Professor Dat during the Spring '11 term at Bilkent University.
 Spring '11
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