prob5 - Mathematical Systems Probability Solutions by...

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Unformatted text preview: Mathematical Systems Probability Solutions by Bracket A First Course in Probability Chapter 5—Problems 2. 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 f ( x ) = Cxe- x/ 2 x > x ≤ what is the probability that the system functions for at least 5 months. Note that we must have R ∞ Cxe- x/ 2 = C [ e- x/ 2 ( x + 2)] ∞ =- 2 C = 1, and thus C =- 1 2 . Then, P X ≥ 5 = R ∞ 5 Cxe- x/ 2 =- 1 2 [ e- x/ 2 ( x + 2)] ∞ 5 = 7 2 e- 5 / 2 . 4. The probability density function of X , the lifetime of a certain type of electronic device (measured in hours), is given by f ( x ) = 10 x 2 x > 10 x ≤ 10 a. Find P X > 20 . R ∞ 20 f ( x ) = 10 R ∞ 20 x- 2 = 10[- 1 x ] ∞ 20 = 1 2 . b. What is the cumulative distribution function of X ? F ( a ) = R a-∞ f ( x ) dx = 10 R a 10 x- 2 dx =- 10[ 1 x ] a 10 = 1- 10 a for a > 10, and 0 otherwise. 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? The probability of any one of the devices lasting for more than 15 hours is 1- F (15) = 2 3 . Thus the probability is ( 6 3 ) ( 2 3 ) 3 ( 1 3 ) 3 ≈ . 219. This assumes that the length of time that any particular device functions is independent from any other device. 6. Compute E [ X ] if X has a density function given by a. f ( x ) = 1 4 xe- x/ 2 x > 0, otherwise. E [ X ] = Z ∞-∞ xf ( x ) dx = Z ∞ 1 4 x 2 e- x/ 2 dx = 1 4 Z ∞ x 2 e- x/ 2 dx = 1 4 [- 2 e- x/ 2 ( x 2 + 4 x + 8)] ∞ = 1 4 (16) = 4 b. f ( x ) = c (1- x 2 )- 1 < x < 1, otherwise. E [ X ] = Z ∞-∞ xf ( x ) dx = Z 1- 1 xc (1- x 2 ) dx =- 1 4 c [(1- x 2 ) 2 ] 1- 1 = 0 1 Mathematical Systems Probability c. f ( x ) = 5 x 2 x > 5, x ≤ 5. E [ X ] = Z ∞-∞ xf ( x ) dx = 5 Z ∞ 5 1 x = 5[ln x ] ∞ 5 = ∞ 11. A point is chosen at random on a line segment of length L . Interpret this statement and find the probability that the ratio of the shorter to the longer segment is less than 1 4 . We may take the line segment to be the interval [0 , L ], and let X denote the point on [0 , L ] that we choose. We’ll assume that the selection of the point is uniformly distributed, and thus f X ( x ) = 1 L for x ∈ [0 , L ] and 0 otherwise. It follows that F X ( p ) = p L for all p ∈ [0 , L ], and 0 or 1 otherwise....
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This note was uploaded on 03/01/2009 for the course PSTAT 120A taught by Professor Mackgalloway during the Spring '08 term at UCSB.

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prob5 - Mathematical Systems Probability Solutions by...

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