744 views
1 votes
1 votes

In this question if we do simply probability calculation then it is 8/20 40%

but when I am appling poisson distribution then it is 40.4%.

why we are getting two different answers??

1 Answer

2 votes
2 votes
$\lambda =\large ( \frac{4}{20} \times 0 ) + (  \frac{3}{20} \times 1)  + ( \frac{5}{20} \times 2 )+ ( \frac{2}{20} \times 3 ) + ( \frac{4}{20} \times 4) + (\frac{5}{20} \times 5  )+ (  \frac{1}{20} \times 6 ) = 2.3  $

$P\{X = i\} =  \Large \frac{e^{\lambda}\lambda^{i}}{i!}$

$P\{X = i + 1\} = P(i) \times  \Large \frac{\lambda}{i+1}$

$P\{X = 0\} =  \Large \frac{e^{-2.3}\times 2.3^{0}}{0!} = 0.100$

$P\{X = 1\} = P(0) \times  \Large \frac{2.3}{0+1} = 0.230$

$P\{X = 2\} = P(1) \times  \Large \frac{2.3}{1+1} = 0.264$

$P\{X > 2\} = 1 - 0.100 - 0.230 - 0.264 = 0.406$

Related questions

1 votes
1 votes
1 answer
2
LRU asked Nov 25, 2022
1,418 views
Let X have a Poisson distribution with parameter λ = 1. What is the probability that X ≥ 2 given that X ≤ 4?
3 votes
3 votes
0 answers
3
Mk Utkarsh asked Aug 31, 2018
907 views
If x has a modified Poisson distribution$P_k = P_r(x = k) =$$\Large \frac{ ( e^m - 1 )^{-1} m^k}{k!}$, $(k = 1,2,3.....)$, then expected value of x is .......
1 votes
1 votes
1 answer
4
saumya mishra asked Aug 20, 2017
614 views
The second moment of a poisson distributed random variable is 2 the mean of the random variable is?