Question
Download Solution PDFConsider a random variable to which a Poisson distribution is best fitted. It happens that \({P(x=1)} = \frac{2}{3}{P({x\;=\;2})}\) on this distribution plot. The variance of this distribution will be
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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:
Given:
\({P_{x\;=\;1}} = \frac{2}{3}{P_{x\;=\;2}}\)
For Poisson distribution:
\({\rm{Probability}},{\rm{\;P\;}}\left( {{\rm{x\;}} = {\rm{\;r}}} \right) = \frac{{{e^{ - λ }} \;{λ ^r}}}{{r!}}\)
\( \Rightarrow \frac{{{e^{ - λ}}{λ^1}\;}}{{1!}} = \frac{2}{3}\frac{{{e^{ - λ}}{λ^2}}}{{2!}}\)
⇒ λ = 3
For Poisson distribution,
Mean = Variance = λ
∴ Variance of this distribution = 3Last updated on Jul 2, 2025
-> ESE Mains 2025 exam date has been released. As per the schedule, UPSC IES Mains exam 2025 will be conducted on August 10.
-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.