In Poisson distribution if P(X=4) = P(X=5) then the parameter of Poisson distribution is:

Option d: 5. Solution: Parameter of a Poisson Distribution given P(X=4) = P(X=5) Let m be the parameter (mean) of the Poisson distribution. The probability mass function is: P(X = x) = -m mxx! Step 1: Equate the probabilities P(X=4) and P(X=5) Given: P(X = 4) = P(X = 5) -m m44! = -m m55! Step 2: Solve for m Since e-m > 0 and m > 0 , we can divide both sides by -m m44! : 1 = /5 m = 5 Therefore, the parameter is 5 . Both Option B and Option D are given as 5 . Since the correct option in the key is listed as Option D, we choose Option D. Hence, Option D is the correct answer.

In Poisson distribution if $P(X=4) = P(X=5)$ then the parameter of Poisson distribution is:

The correct answer is Option d: $5$. **Parameter of a Poisson Distribution given $P(X=4) = P(X=5)$** Let $m$ be the parameter (mean) of the Poisson distribution. The probability mass function is: $$P(X = x) = \\frac{e^{-m} m^x}{x!}$$ **Step 1: Equate the probabilities $P(X=4)$ and $P(X=5)$** Given: $$P(X = 4) = P(X = 5)$$ $$\\frac{e^{-m} m^4}{4!} = \\frac{e^{-m} m^5}{5!}$$ **Step 2: Solve for $m$** Since $e^{-m} > 0$ and $m > 0$, we can divide both sides by $\\frac{e^{-m} m^4}{4!}$: $$1 = \\frac{m}{5}$$ $$m = 5$$ Therefore, the parameter is $5$. Both Option B and Option D are given as $5$. Since the correct option in the key is listed as Option D, we choose Option D. Hence, **Option D** is the correct answer.

Theoretical DistributionsMTP June 24 Series IIQuestion 3980 of 230

All Questions

In Poisson distribution if $\displaystyle P(X=4) = P(X=5)$ then the parameter of Poisson distribution is:

Options

\displaystyle 4

\displaystyle 5

\displaystyle 4

\displaystyle 5

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Correct Answer

✅ Option d — $\displaystyle 5$

All Options:

A $\displaystyle 4$
B $\displaystyle 5$
C $\displaystyle 4$
D $\displaystyle 5$

Detailed Solution & Explanation

**Parameter of a Poisson Distribution given

\displaystyle P(X=4) = P(X=5)

** Let

\displaystyle m

be the parameter (mean) of the Poisson distribution. The probability mass function is:

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

**Step 1: Equate the probabilities

\displaystyle P(X=4)

and

\displaystyle P(X=5)

** Given:

P(X = 4) = P(X = 5)

\\frac{e^{-m} m^4}{4!} = \\frac{e^{-m} m^5}{5!}

**Step 2: Solve for

\displaystyle m

** Since

\displaystyle e^{-m} > 0

and

\displaystyle m > 0

, we can divide both sides by

\displaystyle \\frac{e^{-m} m^4}{4!}

1 = \\frac{m}{5}

m = 5

Therefore, the parameter is

\displaystyle 5

. Both Option B and Option D are given as

\displaystyle 5

. Since the correct option in the key is listed as Option D, we choose Option D. Hence, **Option D** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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In Poisson distribution if P(X=4)=P(X=5)\displaystyle P(X=4) = P(X=5)P(X=4)=P(X=5) then the parameter of Poisson distribution is: