On average, people born in the years 1957 to 1964 held an average of twelve jobs from ages 18 to 52. Estimate the probability that you will hold exactly three jobs in your life. Hint: Model it as a Poisson variable.
My Attempt
Modeling as a Poisson distribution, we have $X\sim Poisson(\lambda)$. From the problem, we are given $\mathbb{E}[X]=12=\lambda$. Then, the probability that you will hold three jobs is $$\mathbb{P}(X=3)=e^{-\lambda}\frac{\lambda^k}{k!}=e^{-12}\frac{12^3}{3!}=0.00177$$
My concern is that the probability is so small, so I am not sure if I did it correctly. It doesn't seem likely that the probability of holding three jobs in your life is only one in one thousand. Can someone verify this or tell me where I went wrong? Thank you.
Yup... a Poisson distribution will have a mode at around its mean and anything besides its mean will be lesser in probability.
This is what the pf looks like for $\text{Poisson}(\lambda=12)$ for the first $21$ values of $x$
To find the mode you could examine the ratio $$\frac{Pr(X=x+1)}{Pr(X=x)}=\frac{e^{-\lambda}\lambda^{x+1}}{(x+1)!}\cdot \frac{x!}{e^{-\lambda}\lambda^x}=\frac{\lambda}{x+1}$$
So the probability is increasing when $x+1<\lambda$ and after a certain point the probability starts decreasing when $x+1>\lambda$ (because $x$ is monotonically increasing). If $\lambda$ is an integer then the modes will be at $\lambda-1$ and $\lambda$ (which happens in the below picture!) and if $\lambda$ is non-integral then the mode happens at $\lfloor \lambda \rfloor$
For $\lambda=12$ there are modes at $11, 12$ and anything in the tails has lesser probability (including $x=3$)