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Test: Poisson Distribution - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test - Test: Poisson Distribution

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Test: Poisson Distribution - Question 1

Let the probability density function of a random variable x be given as

f(x) = ae-2|x|

The value of ‘a’ is __________. 


Detailed Solution for Test: Poisson Distribution - Question 1

Concept:

The probability density function for any f(x) is defined as:

Calculation:

Test: Poisson Distribution - Question 2

If X is a random variable with Mean μ, then the variance of X, denoted by Var (X) is given by:

Detailed Solution for Test: Poisson Distribution - Question 2

Explanation:

If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by:

Var (X) = E[(X - μ)]2, where μ = E(X)

For a discrete random variable X, the variance of X is obtained as follows:

where the sum is taken all over the values of x for which pX(x) > 0. So the variance of X is the weighted average of the squared deviations from the mean μ, where the weights are given by the probability function pX(x) of X.

Important Points

  • The standard deviation of X is defined as the square root of the variance.
  • The variance cannot be negative, because it is an average of squared quantities.
  • Var (X) is often denoted as σ2.

Hence, If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by Var (X) = E[(X - μ)]2

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Test: Poisson Distribution - Question 3

Suppose the probability that a coin toss shows "head" is p, where 0 < p < 1. The coin is tossed repeatedly until the first "head" appears. The expected number of tosses required is

Detailed Solution for Test: Poisson Distribution - Question 3

Let, X, be the number of tosses,

The probability that a coin toss "head" is p, where 0 < p < 1

The coin is tossed repeatedly until the first "head" appears.

Using Binomial expression,

Test: Poisson Distribution - Question 4

A discrete random variable X has the probability functions as:

The value of E(X) is:

Detailed Solution for Test: Poisson Distribution - Question 4

We know that for Fpr random variable X sum of Probaboility = 1

⇒ ∑Pi = 1

⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1

⇒ 26k = 1

⇒ k = 1/26

∴ xf(x) = 103k = 103 × 1/26

∴ E(x) is 103/26

Test: Poisson Distribution - Question 5

Let X be a continuous random variable with PDF

Detailed Solution for Test: Poisson Distribution - Question 5

Test: Poisson Distribution - Question 6

The graph of the probability density function of a random variable X is shown below. What should be the value of λ for this probability density function to be valid?

Detailed Solution for Test: Poisson Distribution - Question 6

Concept:

We know, for valid PDF :

Calculation:

The graph of the pdf is given in triangular form with the base length as 3 and the height as λ.

We know, for valid PDF :

The above is representing the area under the curve between x ϵ [-2, 1].

⇒ 1/2 (base) × Height = 1 

⇒ 1/2 (1- (-2)) × λ = 1 

⇒ λ = 2/3

Test: Poisson Distribution - Question 7

Given a PDF of a random variable X is:

If a = -1 and b = 2, then for c = ½, P(|x| ≤ c) =

Detailed Solution for Test: Poisson Distribution - Question 7

Concept:

  • When we integrate the density function over an interval, we get the probability for that interval, i.e.

P(a ≤ x ≤ b) =

  • Also, for all the values of x,

.

Calculation:

Given: Here, 1/(b - a) = 1/3

Test: Poisson Distribution - Question 8

A man draws 3 balls from a jug containing 5 white balls and 7 black balls. He gets Rs. 20 for each white ball and Rs. 10 for each black ball. What is his expectation?

Detailed Solution for Test: Poisson Distribution - Question 8

3 balls can be drawn in the following ways

Case (i) : Probability of drawing 3 white balls out of 12 balls

Money he gets for drawing 3 white balls

P1 = 3 × 20 = Rs. 60

Case (ii) : Probability of drawing 2 white balls and 1 black ball out of 12 balls

Money he gets for drawing 2 white balls and 1 black ball

P2 = (20 × 2) + (10 × 1) = Rs. 50

Case (iii) : Probability of drawing 1 white ball and 2 black balls out of 12 balls

Money he gets for drawing 3 black balls

P3 = 10 × 3 = Rs. 30

Expectation = Sum of the product of probability and the money he gets for each combination.

Test: Poisson Distribution - Question 9

Let X1 and X2 be two independent exponentially distributed random variables with means 0.5 and 0.25, respectively. Then Y = min (X1, X2) is

Detailed Solution for Test: Poisson Distribution - Question 9

x1 & x2: two independent exponentially distributed random variables with means 0.5 and 0.25.

A continuous random variable x is said to have an exponential (λ) distribution if it has probability density function.

Where, λ > 0, is called the rate of distribution. The mean of the exponential (λ) distribution is calculated using integration by parts as:

Let x1, x2 …..., xn be independent random variables, with xi having exponential (λi) distribution. Then the distribution of min (x1, x2 ……, xn) is exponential (λ1 + λ2 + …. + λn)

Thus, y = min (x1, x2) is exponentially distributed with mean (1/6).

Test: Poisson Distribution - Question 10

If X is a Poisson variate with P(X = 0) = 0.6, then the variance of X is:

Detailed Solution for Test: Poisson Distribution - Question 10

Given

In Poisson distribution

P(X = 0) = 0.6

Formula

Poisson distribution is given by

f(x) = eλx/x!

Calculation

P(X = 0) = eλ0/0!

⇒ 0.6 = e

⇒ 1/eλ = 6/10 = 3/5

⇒ eλ = 5/3

Taking log on both side

⇒ logeeλ = loge(5/3)

∴ λ = Loge(5/3)

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