Denote a number of successes as k k k .
Then probability for Poisson distribution is
P ( k ) = λ k ⋅ e λ k ! P(k)=\dfrac{\lambda^k\cdot e^\lambda}{k!} P ( k ) = k ! λ k ⋅ e λ
Mean and variance are both equal to λ \lambda λ .
So,
mean E [ X ] = λ \bold{E}[X]=\lambda E [ X ] = λ
standard deviation σ = V a r [ X ] = λ \sigma=\sqrt{\bold{Var}[X]}=\sqrt{\lambda} σ = Var [ X ] = λ .
P ( 3 ) = λ 3 ⋅ e λ 3 ! P(3)=\dfrac{\lambda^3\cdot e^\lambda}{3!} P ( 3 ) = 3 ! λ 3 ⋅ e λ
P ( 4 ) = λ 4 ⋅ e λ 4 ! P(4)=\dfrac{\lambda^4\cdot e^\lambda}{4!} P ( 4 ) = 4 ! λ 4 ⋅ e λ
P ( 3 ) = 2 3 P ( 4 ) P(3)=\dfrac{2}{3}P(4) P ( 3 ) = 3 2 P ( 4 )
λ 3 ⋅ e λ 3 ! = 2 3 ⋅ λ 4 ⋅ e λ 4 ! \dfrac{\lambda^3\cdot e^\lambda}{3!}=\dfrac{2}{3}\cdot\dfrac{\lambda^4\cdot e^\lambda}{4!} 3 ! λ 3 ⋅ e λ = 3 2 ⋅ 4 ! λ 4 ⋅ e λ
λ 3 3 ! = 2 3 ⋅ λ 4 4 ! \dfrac{\lambda^3}{3!}=\dfrac{2}{3}\cdot\dfrac{\lambda^4}{4!} 3 ! λ 3 = 3 2 ⋅ 4 ! λ 4
1 3 ! = 2 3 ⋅ λ 4 ! \dfrac{1}{3!}=\dfrac{2}{3}\cdot\dfrac{\lambda}{4!} 3 ! 1 = 3 2 ⋅ 4 ! λ
λ 4 ! = 3 2 ⋅ 1 3 ! \dfrac{\lambda}{4!}=\dfrac{3}{2}\cdot\dfrac{1}{3!} 4 ! λ = 2 3 ⋅ 3 ! 1
λ = 3 2 ⋅ 4 ! 3 ! \lambda=\dfrac{3}{2}\cdot\dfrac{4!}{3!} λ = 2 3 ⋅ 3 ! 4 !
λ = 3 2 ⋅ 3 ! ⋅ 4 3 ! = 3 ⋅ 2 = 6 \lambda=\dfrac{3}{2}\cdot\dfrac{3!\cdot4}{3!}=3\cdot2=6 λ = 2 3 ⋅ 3 ! 3 ! ⋅ 4 = 3 ⋅ 2 = 6
Answer
Mean E [ X ] = 6 \bold{E}[X]=6 E [ X ] = 6
Standard deviation σ = 6 \sigma=\sqrt{6} σ = 6
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