Test: Poisson Distribution - Engineering Mathematics MCQ

Concept:

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

Calculation:

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

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,

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

⇒ ∑P_i = 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

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

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

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.

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 …..., x_n be independent random variables, with x_i having exponential (λ_i) distribution. Then the distribution of min (x₁, x₂ ……, x_n) is exponential (λ₁ + λ₂ + …. + λ_n)

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

Given

In Poisson distribution

P(X = 0) = 0.6

Formula

Poisson distribution is given by

f(x) = e^-λλ^x/x!

Calculation

P(X = 0) = e^-λλ⁰/0!

⇒ 0.6 = e^-λ

⇒ 1/e^λ = 6/10 = 3/5

⇒ e^λ = 5/3

Taking log on both side

⇒ log_ee^λ = log_e(5/3)

∴ λ = Log_e(5/3)