Discussion
Let X be Poisson variate with mean \(\lambda\). Then distribution function of X, evaluated at 1.5 is :
Answer: C
The value is equal to \((\lambda + 1)e^{-\lambda}\).
A Poisson variable takes only integer values, so its distribution function is a step function, and \(F(1.5) = P(X \le 1.5) = P(X \le 1)\):
\[ \begin{aligned} &P(X = 0) + P(X = 1) \\ &= e^{-\lambda} + \lambda e^{-\lambda} \\ &= (\lambda + 1)e^{-\lambda} \end{aligned} \]
It is not 0, since \(P(X = 0) \gt 0\), and it is at least \(e^{-\lambda}\), so B is false. Option D would need it to exceed \(\lambda(\lambda + 1)e^{-\lambda}\), which fails whenever \(\lambda \ge 1\).
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development