31. The probability density function of a signal x[ t] is given by
    

    Px(x) =	1	|x|. e– |x|
    	2

    
    the probability that x > 1 is—

1. 1. 0·247

   
   
   

   2. 0·184

   
   
   

   3. 0·368

   
   
   

   4. 0·493

   
   
   

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1. View Hint View Answer Discuss in Forum

   P(x > 1) =		∞	Px(x).dx =		∞	|x|	e–|x|dx
   		1			1	2

   
   

   =		∞	|x|	e–xdx
   		1	2

   
   

   =

   1

   x

   ∞

   e–xdx –

   d

   .x

   ∞

   e–xdx

   2

   1

   dx

   1

   
   

   =	1		x-x	+		∞	e-x	dx
   	2		- 1			1	1

   
   

   =	1	[– xe–x – e–x]∞1
   	2

   
   = 1/e = 0·368

   Correct Option: C

   P(x > 1) =		∞	Px(x).dx =		∞	|x|	e–|x|dx
   		1			1	2

   
   

   =		∞	|x|	e–xdx
   		1	2

   
   

   =

   1

   x

   ∞

   e–xdx –

   d

   .x

   ∞

   e–xdx

   2

   1

   dx

   1

   
   

   =	1		x-x	+		∞	e-x	dx
   	2		- 1			1	1

   
   

   =	1	[– xe–x – e–x]∞1
   	2

   
   = 1/e = 0·368

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32. The probability density function of a signal x[ t] is given by
    

    Px(x) =	1	|x|. e– |x|
    	2

    
    the probability that – 1 ≤ x ≤ 2

1. 1. 0·74

   
   
   

   2. 0·43

   
   
   

   3. 0·68

   
   
   

   4. 0·39

   
   
   

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1. View Hint View Answer Discuss in Forum

   P(– 1 < x ≤ 2)
   

   =		∞	Px(x)dx =		∞	|x|	e-xdx
   		- ∞			- ∞	2		0

   
   

   =		0	-	x	exdx +		2	x	e-xdx
   		- 1		2			0	2		0

   
   

   =	1			xex -		∞	xexdx		-		xe-x	+		2	xe-xdx
   	2					- 1					- 1			0

   
   

   = 1 -	1	–	3	= 0.42
   	e		e2

   Correct Option: B

   P(– 1 < x ≤ 2)
   

   =		∞	Px(x)dx =		∞	|x|	e-xdx
   		- ∞			- ∞	2		0

   
   

   =		0	-	x	exdx +		2	x	e-xdx
   		- 1		2			0	2		0

   
   

   =	1			xex -		∞	xexdx		-		xe-x	+		2	xe-xdx
   	2					- 1					- 1			0

   
   

   = 1 -	1	–	3	= 0.42
   	e		e2

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33. The probability p(x > 4) is—

1. 1. 0·5

   
   
   

   2. 0·438

   
   
   

   3. 0·948

   
   
   

   4. 0·843

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that general Gaussion probability density function is given by
   

   Px(x) =	1	e – (x – μx)2/2σ2x
   	√2πσx

   
   On comparing with given equation
   

   Px(x) =	1	e – (x – μx)2/2σ2x
   	√2πσx

   
   μx = 4, σ_x = 3
   

   P(x > 4) = Q		x – μx		= Q		4 – 4
   		σx				3

   
   = Q(0) = 0·5

   Correct Option: A

   We know that general Gaussion probability density function is given by
   

   Px(x) =	1	e – (x – μx)2/2σ2x
   	√2πσx

   
   On comparing with given equation
   

   Px(x) =	1	e – (x – μx)2/2σ2x
   	√2πσx

   
   μx = 4, σ_x = 3
   

   P(x > 4) = Q		x – μx		= Q		4 – 4
   		σx				3

   
   = Q(0) = 0·5

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34. The probability p(x > 0) is—

1. 1. – 0·90824

   
   
   

   2. 0·90824

   
   
   

   3. 0·09176

   
   
   

   4. – 0·09176

   
   
   

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1. View Hint View Answer Discuss in Forum

   P(x > 4) = Q		0 - 4		= Q		– 4
   		3				3

   
   

   = 1 – Q		4
   		3

   
   = 1 – 0·09176
   = 0·90824

   Correct Option: B

   P(x > 4) = Q		0 - 4		= Q		– 4
   		3				3

   
   

   = 1 – Q		4
   		3

   
   = 1 – 0·09176
   = 0·90824

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35. The probability p(x > – 2) is—

1. 1. 0·5

   
   
   

   2. 0·97725

   
   
   

   3. 0·90824

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   P(x > – 2) = Q		x – μx
   		σx

   
   

   = Q		– 2 – 4
   		3

   
   = Q (– 2)
   = 1 – Q (2)
   = 1 – 0·02275 = 0·97725

   Correct Option: B

   P(x > – 2) = Q		x – μx
   		σx

   
   

   = Q		– 2 – 4
   		3

   
   = Q (– 2)
   = 1 – Q (2)
   = 1 – 0·02275 = 0·97725

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