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पाईये Continuous Random Variable and Probability Density Function उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Continuous Random Variable and Probability Density Function एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Continuous Random Variable and Probability Density Function MCQ Objective Questions

Top Continuous Random Variable and Probability Density Function MCQ Objective Questions

Continuous Random Variable and Probability Density Function Question 1:

The pdf of a random variable X is expressed by the Quadratic function.

fX(x) = ax(1 - x) for 0 < x < 1.

= 0 Otherwise

The probability P(0.4 < X < 0.9) is __________. [upto 2 decimals]

Answer (Detailed Solution Below) 0.60 - 0.64

Continuous Random Variable and Probability Density Function Question 1 Detailed Solution

Concept:

Area under probability density function pdf = 1

+fx(x)dx=1

P(a < x < b)

=abfx(x)dx

Calculation:

01ax(1x)dx=1

a[01xdx01x2dx]

a[(x22)01(x33)01]

a[1213]

a6=1

a = 6

fx(x) = 6x(1 - x)

Probability:

P(0.4<x<0.9)=0.40.96x(1x)dx=0.62

Continuous Random Variable and Probability Density Function Question 2:

The noise voltage in an electric circuit can be modeled as Gaussain Random variable with mean equal to zero and variance equal to 10-8. The Probability that the noise value exceeds 10-4 is Q(x). The value of x is __________.

Answer (Detailed Solution Below) 1

Continuous Random Variable and Probability Density Function Question 2 Detailed Solution

The random variable X is Gaussian with zero mean

variance = σ2 = 10-8

Q(x) is the probability that a normal Gaussian random variable will obtain a value larger than ‘x’.

Normalisation X=Yuσ

Thus Probability p (X > x) =Q(xσ)

P(x>104)=Q(104104)=Q(1) 

Continuous Random Variable and Probability Density Function Question 3:

If x is a continuous real value random variable defined over the interval (-∞, +∞) and its occurrence is defined by the density function

f(x)=12πbe12(xab)2

Where ‘a’ & ‘b’ are statistical attributes of random variable x. The value of the integral

af(x)dx is ______

Answer (Detailed Solution Below) 0.5

Continuous Random Variable and Probability Density Function Question 3 Detailed Solution

In normal distribution, the area under the normal curve from -∞ to mean =12

Here mean = ‘a’

=af(x)dx=0.5

Continuous Random Variable and Probability Density Function Question 4:

A continuous random variable, X is distributed in interval 0 to 10.

The probability, P(X = 2) is ______.

Answer (Detailed Solution Below) 0

Continuous Random Variable and Probability Density Function Question 4 Detailed Solution

In case of a continuous random variable, the probability at a point = 0

It is only when we integrate the density function over an interval, we get the probability for that interval, i.e.

P(a ≤ x ≤ b) = abfx(x)dx=1

where fx(x) is any given density function.

Continuous Random Variable and Probability Density Function Question 5:

The probability density function curve for a random variable ‘x’ is given below. The value of ‘a’ is ___________.

29.06.2018.012

Answer (Detailed Solution Below) 1.2 - 1.4

Continuous Random Variable and Probability Density Function Question 5 Detailed Solution

The probability density function integration over entire real number is equal to 1

i.e. fx(x)dx=1 

⇒ The area under the curve of pdf is equal to 1

12×3×18+1×14+12×1×(a25)=1716+a0.252=1a0.252=916a=118

= 1.375

Continuous Random Variable and Probability Density Function Question 6:

A probability density function is given by P(X) =π2kex22,<X. The value of k should be

  1. 12π
  2. 2π
  3. 12π
  4. 1π2

Answer (Detailed Solution Below)

Option 2 : 2π

Continuous Random Variable and Probability Density Function Question 6 Detailed Solution

The standard Gaussian pdf is given by, fX(x)=12πex22

Comparing, with the above pdf we have,

π2k=12π

k=12π×2π=2π

Continuous Random Variable and Probability Density Function Question 7:

The probability density function PX(x) of X is shown below:

Gate EC Schlarship Images-Q28

The mean of the variable X is _________

Answer (Detailed Solution Below) 1.1 - 1.125

Continuous Random Variable and Probability Density Function Question 7 Detailed Solution

For PX(x) to be valid pdf, PX(x)dx=1 

12×1+38×1+a×1=1a=11238=8438=18a=1/8

Now, mean m of X is xPX(x)dx

m=0112xdx+1238xdx+2318xdx=14+916+516m=4+9+516=1816=1.125

Continuous Random Variable and Probability Density Function Question 8:

The pdf of a Gaussian random variable X is given by

pX(x)=14πexp[(x3)216]

The probability of the event X=3 is _______.

Answer (Detailed Solution Below) 0

Continuous Random Variable and Probability Density Function Question 8 Detailed Solution

For continuous random variable distributions, the probability is defined only for a range of values and not any discrete point. The probability of any discrete point is always zero.

Continuous Random Variable and Probability Density Function Question 9:

Comprehension:

Consider a baseband binary PAM receiver shown below. The addition channel noise n(t) is white with power spectral density. The low pass filter is ideal with unity gain and cut off frequency 1 MHz. Let Yk represent the random variable y(tk)

Yk=Nk, if transmitted bit bk=0.

Yk=a+Nk, if transmitted bit bk=1.

Where, Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5αeα|n|. (This has mean zero and variance 2α2). Assume transmitted bits to be equiprobable and threshold z is set to a2=106V.

4

The probability of bit error is

  1. 0.5×e35
  2. 0.5×e5
  3. 0.5×e7
  4. 0.5×e10

Answer (Detailed Solution Below)

Option 4 : 0.5×e10

Continuous Random Variable and Probability Density Function Question 9 Detailed Solution

O/p Noise Power = o/p PSD × B.W

= 10-20 × 2 × 106

= 2 × 10-14 W

Since mean square value = Power

 2α2=2×1014α=107

When a 1 is transmitted:

Yk = a + Nk

Threshold

Z= a/2= 10-6

⇒ a = 2 × 10-6

For error to occur, Yk < 10-6

2 × 10-6 + Nk < 10-6

Nk < - 10-6

P(01)=106P(n)dn

106(0.5)α.eαn.dn,withα=107

= 0.5 × e-10

When a ‘0’ is Transmitted:

Yk = Nk

For error to occur, Yk > 10-6

P(10)=106P(n)dn=0.5×e10

Since both bits are equiprobable 

Pe=12[P(0/1)+P(1/0)]

0.5 x e-10.

Continuous Random Variable and Probability Density Function Question 10:

Comprehension:

Consider a baseband binary PAM receiver shown below. The addition channel noise n(t) is white with power spectral density. The low pass filter is ideal with unity gain and cut off frequency 1 MHz. Let Yk represent the random variable y(tk)

Yk=Nk, if transmitted bit bk=0.

Yk=a+Nk, if transmitted bit bk=1.

Where, Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5αeα|n|. (This has mean zero and variance 2α2). Assume transmitted bits to be equiprobable and threshold z is set to a2=106V.

4

The value of the parameter α(inV1) is 

  1. 1010
  2. 107
  3. 1.414×1010
  4. 2×1026

Answer (Detailed Solution Below)

Option 2 : 107

Continuous Random Variable and Probability Density Function Question 10 Detailed Solution

O/p Noise Power = o/p PSD × B.W

= 10-20 × 2 × 106

= 2 × 10-14 W

Since mean square value = Power

 2α2=2×1014

α=107

When a 1 is transmitted:

Yk = a + Nk

Threshold

Z= a/2= 10-6

⇒ a = 2 × 10-6

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