Probability and Random Variable MCQ || Random Variables Questions and Answers

Answer.1. 0.25

Explanation

If X be a random variable with a finite number of outcomes x₁, x_2, x_{3 ….} occurring with probabilities p₁, p₂, p₃……. respectively, the expectation of X is defined as:

$E\left[ x \right] = \mathop \sum \limits_{i = 1}^k {x_i} \cdot {p_i} = {x_1}{p_1} + {x_2}{p_2} \ldots$

Application:

It is given that the two sides of a fair coin are labelled as 0 and 1 and the coin is tossed two times independently.

∴ The total possible outcomes can be:

O = {(1,1), (1,0), (0,1), (0,0)}

The random variable X is defined as:

X = min (M, N)

So,

X₁ = min {(1, 1)} = 1

X₂ = min {(1, 0)} = 0

X₃ = min {(0, 1)} = 0

X₄ = min {(0, 0)} = 0

The probability of occurrence of ‘1’ will be:

P(1) = 1/4

And the probability of occurrence of ‘0’ will be:

P(0) = 3/4

We know that:

$\begin{array}{l} E\left[ x \right] = \mathop \sum \limits_{i = 1}^k {x_i} \cdot {P_i}\\ \\ E\left[ x \right] = 0 \times \frac{3}{4} + 1 \times \frac{1}{4} \end{array}$

E[x] = 0.25

Answer.1. Gaussian distribution

Explanation

Gaussian distribution is often used in statistics and is often used in the natural and social sciences to represent real-valued random variables whose distributions are not known. Eg-Thermal noise.

Normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In graph form, normal distribution will appear as a bell curve.

Answer.3. 300 Hz

Explanation

If X be a random variable with a finite number of outcomes x₁, x_2, x_{3 ….} occurring with probabilities p₁, p₂, p₃……. respectively, the expectation of X is defined as:

$E\left[ x \right] = \mathop \sum \limits_{i = 1}^k {x_i} \cdot {p_i} = {x_1}{p_1} + {x_2}{p_2} \ldots$

(n) is proportional to n where n= 1,2,3,…6 is random variable.

P(n) = kn

P(1) + P(2)….P (6) = 1

K(1 + 2 + 3 + 4 + 5 + 6) = 1

K = 1/21

P(4) = 4K

P(4) = 4/21

Answer.2. Rayleigh distribution

Explanation

Rayleigh distribution is used in mobile computing for flat fading channels. The Rayleigh distribution is a continuous probability distribution named after the English Lord Rayleigh. It is a special case of the Weibull distribution with a scale parameter of 2.

Rayleigh fading is caused by multipath reception. The mobile antenna receives a large number, say N, reflected and scattered waves. Because of wave cancellation effects, the instantaneous received power seen by a moving antenna becomes a random variable, dependent on the location of the antenna.

Answer.3. Poisson distribution

Explanation

• Poisson distribution is a theoretical discrete probability distribution that is very useful in situations where discrete events occur in a continuous manner.
• A Poisson distribution, named after French mathematician Siméon Denis Poisson, can be used to estimate how many times an event is likely to occur within “X” periods of time.
• If an event happens independently and randomly over time, and the mean rate of occurrence is constant over time, then the number of occurrences in a fixed amount of time will follow the Poisson distribution.

Answer.3. Poisson distribution

Explanation

In uniform distribution, every outcome is equally likely to occur and is applied for a random experiment with equal chances of occurrences of every possible outcome.

Answer.3. 0.4

Explanation

P (0.5 < x < 5) = Integrating f (x) from

0.5 to 5 by splitting in 3 parts that is from 0.5 to 1

and from 1 to 4 and 4 to 5 we get

P (0.5 < x < 5) = 0.1 + 0.3 + 0

P (0.5 < x < 5) = 0.4.

Answer.1. $\sigma _Y^2 = 4\sigma _X^2$

Explanation

Variance of a random variable ‘y’ is given by:

Var[y] = E[y²] – E²[y]

Properties of mean:

1) E[K] = K, Where K is some constant

2) E[c X] = c. E[X], Where c is some constant

3) E[a X + b] = a E[X] + b, Where a and b are constants

4) E[X + Y] = E[X] + E[Y]

Application:

Variance of y = E[(2x + 3)²] – (E[2x + 3])²

= E[4x² + 12x + 9] – (E[2x + 3])²

= 4E[x²] + 12E[x] + 9 – (E[2x] + E[3])²

= 4E[x2] + 12E[x] + 9 – (2E[x] + 3)2

= 4E[x2] + 12E[x] + 9 – (4E²[x] + 9 + 12E[X])

= 4E[x2] + 12E[x] + 9 – 4E2[x] – 9 – 12E[X]

= 4E[x2] – 4E2[x]

= 4[E[x²] – E²[x]]

This can be written as:

= 4 (variance of x), i.e.

The variance of y = 4 times the variance of x

$\sigma _Y^2 = 4\sigma _X^2$

Answer.1. 0

Explanation

Two random variables are said to be uncorrelated if their covariance is zero.

Correlation Coefficient (ρ) of two Random Variables x and y is given by

${\rho _{xy}} = \frac{{{C_{ov}}\left[ {x,\:y} \right]}}{{\sqrt {{V_{ar}}\left[ x \right]{v_{ar}}\left[ y \right]} }}$

Where,

C_ov [x, y] = E [X, Y] – E [X] E[Y]  —(2)

Analysis:

Given X & Y are totally uncorrelated, i.e.

E[X Y] = E[X] E[Y]

Substituting this is Equation (2), we get:

C_ov [X, Y] = E[X Y] – E [X] E [Y] = 0

∴ C_ov [X, Y] = 0

Hence from eqn (1), ρ_xy = 0

Answer.1. 25.79

Explanation

The mean of 5 innings is

(50 + 70 + 82 + 93 + 20) ÷ 5 = 63

S.D = [¹⁄_n (x(n) − mean)²]^0.5

S.D = 25.79