Poisson Distribution when only given using mean

I'm doing the following homework problem and am unsure of whether or not my answers are correct. This is my first time working with Poisson distribution and I want to make sure I am doing it correctly.

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with mean $u = 20$. What is the probability that the number of drivers will

a. Be at most 10?

b. Exceed 20?

c. Be between 10 and 20, inclusive? Be strictly between 10 and 20?

d. Be within 2 standard deviations of the mean value?

I'm pretty much just trying to follow the formula that I was given for Poisson distribution and have the following answers:

a. $P(x\le 10) = \sum{0\to10}\frac {e^{20} \times 20^x}{x!}$

b. $P(x>20) = \frac {e^{20} \times 20^(20)}{20!}$

c. $P(10 \le x \le 20) = \sum_{10\to20}\frac {e^{20} \times 20^x}{x!}$

$P(10 < x <20) = \sum_{10\to20}\frac {e^{20} \times 20^x}{x!}$

d. not sure still

The Poisson with parameter $\lambda$ has mean $\lambda$ (and variance $\lambda$).

So if you know the mean, you know the parameter.

For the first question, the required probability is $$\sum_{k=0}^{10} e^{-20} \frac{20^k}{k!}.$$ Note that it is $e^{-20}$, not the $e^{20}$ of the post. The same mistake recurs.

For the probability that $X\gt 20$, we could sum from $21$ to $\infty$. But that's a long sum! To evaluate it, calculate the probability that $X\le 20$, and subtract this from $1$.

For c), your sum (apart from the fact that we need $e^{-20}$) is correct. For strictly between, sum from $11$ to $19$.

For d), the standard deviation is $\sqrt{20}$. So $2$ standard deviations is about $8.944$. Find the integers that are within $8.944$ of $20$, and take the sum of the Poisson probabilities in that range.

• So for part d are you saying just to use the formula for Poisson going from 9 to 20? I'm not sure if I completely understood what you meant Mar 14, 2014 at 18:12
• Yea this explanation for D is very unclear. Mar 25, 2015 at 19:15
• The Poisson parameter $\lambda$ has variance $\lambda$, so SD $\sqrt{\lambda}$, here $\sqrt{20}$. So $2$ SD is $2\sqrt{20}$, about $8.944$. That puts us in the range $11.05$ to $28.944$. Now find the sum of the Poisson probabilities $e^{-20}\frac{20^k}{k!}$, $k=12$ to $28$, though a case might be made for $k=11$ to $29$. Alternately, we could treat the Poisson with parameter $20$ as "close enough" to a normal, and use the normal approximation. The probability of being within $2$ SD units, for the normal, is very close to $0.95$, so for this Poisson we would expect roughly $0.95$. Mar 25, 2015 at 19:59

Not quite

(a) You have a missing sign in the exponential, so $$P(X \le 10) = \sum_{x=0}^{10} e^{-20}\frac{20^x}{x!}$$

(b) You could do a sum from $21$ to infinity but it is probably easier to consider $P(X \gt 20) = 1- P(X \le 20)$ and adapt the answer to (a)

(c) again you have a missing sign in the exponential, and for the strictly part the $\le$ have to become $\lt$.

(d) as a hint, the variance of a Poisson distribution is equal to its mean, and the standard deviation is the square root of the variance.