Answer:
1.3125
Step-by-step explanation:
Given that our random variable [tex]X[/tex] follows a Poisson distribution[tex]P(X=k)=\frac{\lambda^k e^-^p}{k!} \ \ \ \ \ \ \ \ p=\lambda[/tex]
Evaluate the formula at [tex]k=2,3:[/tex]
[tex]P(X=2)=0.5\lambda^2e^{-\lambda}\\\\P(X=3)=\frac{1}{6}\lambda^2e^{-\lambda}[/tex]
#since
[tex]4P(X=2)=P(X=3);\\\\0.5\lambda^2e^{-\lambda}=4\times\frac{1}{6}\lambda^3e^{-\lambda}\\\\0.5\lambda^2=\frac{2}{3}\lambda^3\\\\0.75=\lambda[/tex]
The mean and variance of the Poisson distributed random variable is equal to [tex]\lambda[/tex]:
[tex]\mu=\lambda=0.75\\\sigma ^2=\lambda=0.75[/tex]
#By property variance:
[tex]\sigma ^2=V(X)=E(X^2)-(E(X))^2=E(X^2)\\\\E(X^2)=\sigma^2+\mu^2=0.75+0.75^2=1.3125[/tex]
The expectation is 1.3125
