Poisson Distribution MCQ Quiz in తెలుగు - Objective Question with Answer for Poisson Distribution - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Poisson Distribution: A random variable X, taking on one of the values 0,1,2,..... is said to be a Poisson random variable with parameter λ for λ > 0, Probability is:

P (X = x) = \(\frac{e^{-λ}λ^x}{x!}\)

Calculation:

Given:

Prob(X = 1) = Prob(X = 2)

e = 2.718

\(\frac{e^{-λ}λ^1}{1!} = \frac{e^{-λ}λ^2}{2!}\)

λ = λ²/2

(λ² - 2λ) = 0 ⇒ λ (λ - 2) = 0

Since λ is always greater than zero, hence

λ - 2 = 0 ⇒ λ = 2

Prob(X = 3) = \(\frac{e^{-λ}λ^3}{3!} = \frac{e^{-2}.\ 2^3}{2\ \times\ 3}\)

Prob(X = 3) = \( \frac{ 8}{6. e^2}\) = 0.18

Additional InformationFor Poisson distribution :

Mean = E(x) = λ

Variance = V(x) = λ 

Thus, here λ is an average number of occurrences of events in an observation period Δt.

So, λ = α.Δt

where: α = number of occurrences of event per unit time.

Concept:

The probability density function of Poisson’s distribution is

\(p\left( x \right)=\frac{{{e}^{-\lambda }}\cdot {{\lambda }^{x}}}{x!}\)

Where

λ = mean = np

n = number of total outcomes

p = probability of success

Note: This distribution uses where the probability of success i.e. p is very small.

Calculation:

Poisson (2) represents that the distribution is Poisson with λ = 2.

Now, the distribution function \(= \frac{{{e^{ - 2}}{2^x}}}{{x!}}\)

\(E\left( {\frac{1}{{1 + x}}} \right) = \mathop \sum \limits_{x = 0}^\infty \frac{1}{{1 + x}}\left( {\frac{{{e^{ - 2}}{2^x}}}{{x!}}} \right)\)

\(= \mathop \sum \limits_{x = 0}^\infty \frac{{{e^{ - 2}}{2^x}}}{{\left( {x + 1} \right)!}}\)

\(= \frac{{{e^{ - 2}}}}{2}\mathop \sum \limits_{x = 0}^\infty \frac{{{2^{\left( {x + 1} \right)}}}}{{\left( {x + 1} \right)!}}\)

\(= \frac{{{e^{ - 2}}}}{2}\left( {\frac{2}{{1!}} + \frac{{{2^2}}}{{2!}} + \frac{{{2^3}}}{{3!}} + \ldots } \right)\)

\( = \frac{{{e^{ - 2}}}}{2}\left( {{e^2} - 1} \right)\)

\(= \frac{1}{2}\left( {1 - {e^{ - 2}}} \right)\)

The correct answer is - mean = variance

Key Points

• Poisson Distribution
  • The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, provided the events occur with a constant rate and are independent of each other.
• Mean and Variance
  • In a Poisson distribution, the mean (denoted by λ) and the variance are always equal.
  • This is a unique property of the Poisson distribution, distinguishing it from other probability distributions like the normal or binomial distributions.
• Correct Answer Justification
  • Option 1 (mean > variance) is incorrect because in a Poisson distribution, the mean is always equal to the variance, not greater.
  • Option 3 (mean < variance) is incorrect for the same reason as above.
  • Option 4 (mean = standard deviation) is incorrect because the standard deviation is the square root of the variance, and it is not equal to the mean unless the mean is 1.
  • Option 2 (mean = variance) is always true for a Poisson distribution, making it the correct answer.

Additional Information

• Applications of Poisson Distribution
  • Used to model the number of events occurring in a fixed interval of time or space, such as:
    • Number of phone calls received by a call center in an hour.
    • Number of decay events per second from a radioactive source.
    • Number of customer arrivals at a service point in a given time period.
• Key Properties of the Poisson Distribution
  • The events are independent; the occurrence of one event does not affect the probability of another event occurring.
  • Events occur at a constant average rate (λ).
  • The probability of more than one event occurring in an infinitesimally small time interval is negligible.
• Formula for Poisson Probability
  • The probability of observing k events in a fixed interval is given by:

    P(X = k) = (λ^k * e^-λ) / k!

  • Here, λ is the mean (and variance) of the distribution, k is the number of occurrences, and e is Euler's number (approximately 2.718).

Concept:

The probability density function of Poisson’s distribution is

\(p\left( x \right)=\frac{{{e}^{-\lambda }}\cdot {{\lambda }^{x}}}{x!}\)

Where

λ = mean = np

n = number of total outcomes

p = probability of success

Note: This distribution uses where the probability of success i.e. p is very small.

Calculation:

Probability (p) = 0.1

λ = mean = 10 × 0.1 = 1

Probability that exactly two will be defective

\( p\left( x \right)=\frac{{{e}^{-1}}\cdot {{\left( 1 \right)}^{2}}}{2!}\)

= 0.184 ≈ 0.2

Let x be the number of accidents in one week.

P(x < 2) = P(x = 0) + P(x = 1)

\(= {e^{ - 0.5}} + \frac{{{e^{ - 0.5}}\; \times \;0.5}}{{1!}}\)

= 0.9098

\(P\left( {x = 2} \right) = \frac{{{e^{ - 0.5}}\; \times \;{{\left( {0.5} \right)}^2}}}{{2!}}\)

= 0.0758

P(x > 2) = 1 – P(x ≤ 2)

= 1 – (0.9098 + 0.0758) = 0.0144

Probability that there will be zero accidents in 3 week period = [P(x = 0)]³

= (e^-0.5)³

= 0.223

Concept:

In case of Poisson distribution, mean and variance are same.

Calculation:

Given, mean = 7

E[(Y + 3)2] = E [Y2 + 6Y + 9] = E[Y2] + E[6Y] + E[9]

Variance = E[Y2] - (E[Y])2

As, mean = variance = 7

Mean = E[Y]

7 = E[Y2] -  (7)2

7 = E[Y2] - 49

E[Y2] = 56

Concept:

It follows a Poisson distribution as the probability of occurrence is very small.

\({\rm{Probability}},{\rm{\;P\;}}\left( {{\rm{x\;}} = {\rm{\;r}}} \right) = \frac{{{e^{ - λ }} \;{λ ^r}}}{{r!}}\)

where mean E(x) = variance Var (x) = λ and

Standard deviation (σ) = \(\sqrt{λ}\)

Calculation:

For Poisson distribution:

\({\rm{Probability}},{\rm{\;P\;}}\left( {{\rm{x\;}} = {\rm{\;r}}} \right) = \frac{{{e^{ - λ }} \;{λ ^r}}}{{r!}}\)

According to question

P (1) =  P (5)

⇒ \(\frac{{{e^{ - λ}}{λ^1}\;}}{{1!}} = (\frac{{{e^{ - λ}}{λ^5}}}{{5!}})\)

⇒ λ⁴ = 120

⇒ λ = 3.31

Since, for Poisson distribution, 

Mean = Variance = λ 

Hence, mean = 3.31

\(P(X) = \frac{{{m^X}{e^{ - m}}}}{{X!}}\)

\(\frac{{{m^2}{e^{ - m}}}}{{2!}} = \frac{{9{m^4}{e^{ - m}}}}{{4!}} + \frac{{90{m^6}{e^{ - m}}}}{{6!}}\)

\({m^4} + 3{m^2} - 4 = 0\)

Let \({m^2} = t\)

Now the equation will become \({t^2} + 3t - 4 = 0\)

\(\begin{array}{l} \Rightarrow t = - 4,\;1\\ \Rightarrow {m^2} = - 4,1\\ \Rightarrow m = \pm 2i, \pm 1 \end{array}\)

As we know mean of Poisson variate is always non negative and real. hence m = 1 is possible but m = -1 is not possible. hence option 4 is correct.

Answer: Option 4

Concept: 

Poisson Distribution: 

The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event.

Notation: 

X ~ P(λ)  /where λ is mean

Formulas:

P(X=x) = \({e^{ - λ }}\frac{{{λ^x}}}{{x!}}\)

Calculation:

 It is given that we need to find P(|X- 3| < 1),

after simplifying we get P( 2 < X < 4 )

So between 2 and 4, only one integer value possible is 3.

we get

 P(|X- 3| < 1) = \(\dfrac{9}{2} e^{-3}\)

Hence Option 1 is the correct answer.

Concept:

A Poisson process is a stochastic process that counts the number of events and time points at which these events occur in a given time interval. Formula for poisson distribution:

\({\rm{P}}\left( {{\rm{x}},{\rm{\lambda }}} \right) = {\rm{}}\frac{{{{\rm{e}}^{ - {\rm{\lambda }}}}{{\rm{\lambda }}^{\rm{x}}}}}{{{\rm{x}}!}}{\rm{\;for\;x}} = 0,1,2, \ldots ..\)

Explanation:

30 requests are sent in 1 hour.

i.e. in 60 min, 30 requests are sent

in 40 min, number of requests sent are 20.

λ = 20.

x = 0

\(P\left( {0,20} \right) = \;\frac{{{e^{ - 20}}{{20}^0}}}{{0!}} = \;{e^{ - 20}}\)