Tut6 - ECE 316 for the week of June 22-26 2009 Problem 1...

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Unformatted text preview: ECE 316 for the week of June 22-26, 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.

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