# Thread: If the random variable X follows Poisson distribution such that

1. ## If the random variable X follows Poisson distribution such that

If a random variable X follows Poisson distribution such that

P(X=1)=P(X=2)

Compute
i) the mean of distibution
ii) P(X=0)
iii) standard deviation of the distribution

Solution

Poisson distribution $p(x, \lambda) = \frac{\lambda^x}{x!} . e^{- \lambda}$

since $P(X=1)=P(X=2) \Rightarrow P(X=1, \lambda)=P(X=2, \lambda) \Rightarrow \frac{\lambda^1}{1!} . e^{- \lambda} = \frac{\lambda^2}{2!} . e^{- \lambda}$

$\lambda = 2$ .................Am i correct ??????

2. looks good, the mean is lambda and the standard deviation is the square root of lambda

3. Originally Posted by Athena
If a random variable X follows Poisson distribution such that

P(X=1)=P(X=2)

Compute
i) the mean of distibution
ii) P(X=0)
iii) standard deviation of the distribution

Solution

Poisson distribution $p(x, \lambda) = \frac{\lambda^x}{x!} . e^{- \lambda}$

since $P(X=1)=P(X=2) \Rightarrow P(X=1, \lambda)=P(X=2, \lambda) \Rightarrow \frac{\lambda^1}{1!} . e^{- \lambda} = \frac{\lambda^2}{2!} . e^{- \lambda}$

$\lambda = 2$ .................Am i correct ??????