Cars pass through a road junction according to Poisson distribution.An average of 7 cars per minute pass through this junction.
What is the expected number of cars passing through in 3 minutes? My answer is $3 \times 7=21$
What is the probability that exactly the expected number pass through the certain 3 minute period? Do I just use the Poisson distribution formula and substitute mean $=x=21$?
So is my equation below correct?$$\frac{ (exp-21)×21^{21}}{21!}$$
On
Yes, you should indeed use the Poisson distribution formula, with $\mu=21$, as the problem states and $x=21$: $$P(x,\mu) = \frac{e^{-\mu }\cdot \mu^{x}}{x!} = \frac{1}{e^{\mu}}\frac{\mu^x}{x!}$$ $$\implies P(x,\mu) = \frac{\mu^x}{e^{\mu}x!}$$ $$P(x, \mu) = e^{-21}\frac{21^{21}}{21!}$$ $$P(x, \mu) = \frac{1}{e^{21}}\cdot \frac{21^{21}}{21!}$$ $$P(x, \mu) = \frac{21^{21}}{e^{21}\cdot 21!}$$
Through calculus, substituting $e$ with $2.71828$, we get:
$$P(x,\mu)\approx 0.08671$$
$e$: A constant equal to approximately $2.71828$.
$\mu$: The mean number of successes that occur in a specified region.
$x$: The actual number of successes that occur in a specified region.
$P(x; \mu)$: The Poisson probability that exactly $x$ successes occur in a Poisson experiment, when the mean number of successes is $\mu$.
$ e = e$
$\mu = 21$
$x = 21$
$$P(x;\mu) = \frac{e^{-\mu}\times \mu^x}{x!}=\frac{21^{21} \times e^{21}}{21!}$$
So your solution is $$\frac{21^{21} \times e^{-21}}{21!}$$