1. The two sides of a fair coin are labeled as 0 and 1. The coin is tossed two times independently. Let M and N denote the labels corresponding to the outcomes of those tosses. For a random variable X, defined as X = min (M, N), the expected value E(X) (rounded off to two decimal places) is _______

0.25

0.50

0.75

1

Answer.1. 0.25

Explanation

If X be a random variable with a finite number of outcomes x_{1}, x_{2, }x_{3 …. }occurring with probabilities p_{1}, p_{2}, p_{3}……. respectively, the expectation of X is defined as:

2. ________ 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.

Gaussian distribution

Rayleigh distribution

Poisson distribution

Uniform distribution

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.

3. Consider a dice with the property that probability of a face with n dots showing up is proportional to n. The probability of face showing 4 dots is?

1/7

5/42

1/21

4/21

Answer.3. 300 Hz

Explanation

If X be a random variable with a finite number of outcomes x_{1}, x_{2, }x_{3 …. }occurring with probabilities p_{1}, p_{2}, p_{3}……. respectively, the expectation of X is defined as:

(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

4. ________ is used in mobile computing for flat fading channels.

Gaussian distribution

Rayleigh distribution

Poisson distribution

Uniform distribution

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.

5. _________ useful in situations where discrete events occur in a continuous manner.

Gaussian distribution

Rayleigh distribution

Poisson distribution

Uniform distribution

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.

6. In ________ every outcome is equally likely to occur.

Gaussian distribution

Rayleigh distribution

Poisson distribution

Uniform 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.

7. Let X be a random variable with probability distribution function f (x)=0.2 for |x|<1

= 0.1 for 1 < |x| < 4
= 0 otherwise
The probability P (0.5 < x < 5) is _____

0.3

0.5

0.4

0.8

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.

8. Two continuous random variables X and Y are related as Y = 2X + 3. Let $\sigma _X^2$ and $\sigma _Y^2$ denote the variances of X and Y, respectively. The variances are related as

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

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

$\sigma _Y^2 = 25\sigma _X^2$

$\sigma _Y^2 = 5\sigma _X^2$

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

Explanation

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

Var[y] = E[y^{2}] – E^{2}[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[4x^{2} + 12x + 9] – (E[2x + 3])^{2}

= 4E[x^{2}] + 12E[x] + 9 – (E[2x] + E[3])^{2}

= 4E[x2] + 12E[x] + 9 – (2E[x] + 3)2

= 4E[x2] + 12E[x] + 9 – (4E^{2}[x] + 9 + 12E[X])

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

= 4E[x2] – 4E2[x]

= 4[E[x^{2}] – E^{2}[x]]

This can be written as:

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

The variance of y = 4 times the variance of x

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

9. The two random variable X and Y are uncorrelated if and only if their covariance is

0

1

-1

infinity

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

10. Runs scored by batsman in 5 one-day matches are 50, 70, 82, 93, and 20. The standard deviation is ______