**Method 0**: The lazy statistician.

Note that for $y \neq 0$ we have $f(y) = (1-\pi) p_y$ where $p_y$ is the probability that a Poisson random variable takes value $y$. Since the term corresponding to $y = 0$ does not affect the expected value, our knowledge of the Poisson and the linearity of expectation immediately tells us that
$$
\mu = (1-\pi) \lambda
$$
and
$$
\mathbb E Y^2 = (1-\pi) (\lambda^2 + \lambda) \> .
$$

A little algebra and the identity $\mathrm{Var}(Y) = \mathbb E Y^2 - \mu^2$ yields the result.

**Method 1**: A probabilistic argument.

It's often helpful to have a simple probabilistic model for how a distribution arises. Let $Z \sim \mathrm{Ber}(1-\pi)$ and $Y \sim \mathrm{Poi}(\lambda)$ be independent random variables. Define
$$
X = Z \cdot Y \>.
$$
Then, it is easy to see that $X$ has the desired distribution $f$. To check this, note that $\renewcommand{\Pr}{\mathbb P}\Pr(X = 0) = \Pr(Z=0) + \Pr(Z=1, Y=0) = \pi + (1-\pi) e^{-\lambda}$ by independence. Similarly $\Pr(X = k) = \Pr(Z=1, Y=k)$ for $k \neq 0$.

From this, the rest is easy, since by the independence of $Z$ and $Y$,
$$
\mu = \mathbb E X = \mathbb E Z Y = (\mathbb E Z) (\mathbb E Y) = (1-\pi)\lambda \>,
$$
and,
$$
\mathrm{Var}(X) = \mathbb E X^2 - \mu^2 = (\mathbb E Z)(\mathbb E Y^2) - \mu^2 = (1-\pi)(\lambda^2 + \lambda) - \mu^2 = \mu + \frac{\pi}{1-\pi}\mu^2 \> .
$$

**Method 2**: Direct calculation.

The mean is easily obtained by a slight trick of pulling one $\lambda$ out and rewriting the limits of the sum.
$$
\mu = \sum_{k=1}^\infty (1-\pi) k e^{-\lambda} \frac{\lambda^k}{k!} = (1-\pi) \lambda e^{-\lambda} \sum_{j=0}^\infty \frac{\lambda^j}{j!} = (1-\pi) \lambda \> .
$$

A similar trick works for the second moment:
$$
\mathbb E X^2 = (1-\pi) \sum_{k=1}^\infty k^2 e^{-\lambda} \frac{\lambda^k}{k!} = (1-\pi)\lambda e^{-\lambda} \sum_{j=0}^\infty (j+1) \frac{\lambda^j}{j!} = (1-\pi)(\lambda^2 + \lambda) \>,
$$
from which point we can proceed with the algebra as in the first method.

**Addendum**: This details a couple tricks used in the calculations above.

First recall that $\sum_{k=0}^\infty \frac{\lambda^k}{k!} = e^\lambda$.

Second, note that
$$
\sum_{k=0}^\infty k \frac{\lambda^k}{k!} = \sum_{k=1}^\infty k \frac{\lambda^k}{k!} = \sum_{k=1}^\infty \frac{\lambda^k}{(k-1)!} = \sum_{k=1}^\infty \frac{\lambda \cdot \lambda^{k-1}}{(k-1)!} = \lambda \sum_{j=0}^\infty \frac{\lambda^j}{j!} = \lambda e^{\lambda} \>,
$$
where the substitution $j = k-1$ was made in the second-to-last step.

In general, for the Poisson, it is easy to calculate the factorial moments $\mathbb E X^{(n)} = \mathbb E X(X-1)(X-2)\cdots(X-n+1)$ since
$$
e^\lambda \mathbb E X^{(n)} = \sum_{k=n}^\infty k(k-1)\cdots(k-n+1) \frac{\lambda^k}{k!} = \sum_{k=n}^\infty \frac{\lambda^n \lambda^{k-n}}{(k-n)!} = \lambda^n \sum_{j=0}^\infty \frac{\lambda^j}{j!} = \lambda^n e^\lambda \>,
$$
so $\mathbb E X^{(n)} = \lambda^n$. We get to "skip" to the $n$th index for the start of the sum in the first equality since for any $0 \leq k < n$, $k(k-1)\cdots(k-n+1) = 0$ since exactly one term in the product is zero.

Let the first (two point) random variable be $A$ and the second be $B.$ Let $q$ be the probability that $B$ takes on the value $0.$ Instead of $r$ as in your problem statement, we equivalently let the probability that $B$ is $z$ be $0.3-q.$ Equating the means, we have $$0.3x+0.7(6000) = (0.3-q)z+0.7(6000),$$

so $$x = {z \left( 1-{q \over 0.3} \right) }$$

If the variances are identical, then the second moments about the origin are the same also. This gives us

$$E[A^2]= 0.3x^2 + 0.7 (6000)^2 = (0.3-q)z^2 + 0.7(6000)^2$$

Plugging in our expression for $x,$ this leads to
$$0.3 \left[z \left( 1 - {q \over 0.3} \right) \right]^2 = 0.3z^2 -qz^2$$ Solving this for $q$ gives the two solutions $q=0$ or $q=0.3,$ both of which imply that $A$ and $B$ have the same two-point distribution.

So your problem as posed does not have a solution in terms of your definition of an acceptable answer.

## Best Answer

I'll assume you only want samples consisting of non-negative integers (otherwise its obviously not Poisson). Indeed, without those restrictions, it's pretty trivial - though largely meaningless, because the variance-ratio changes as you change scale. (It makes more sense with counts though.)

I'll also assume you want the $n-1$ form of the sample variance (the unbiased form).

Some examples:

These took me a few minutes to construct.

It doesn't really have much significance other than giving a lack of evidence against the Poisson (at least on the basis of the ratio of variance to mean). In no way does it tell you that the data

isPoisson.Edit: For example, here's a sample that is pretty obviously

notPoisson (in the sense that the chance that you could end up with a sample like that from a Poisson isreaallysmall):For starters, all the values are even!

Here's another that's pretty clearly not Poisson:

Edit: here's a biggish sample that's not so plainly inconsistent with Poisson:

... though it's a bit too kurtotic to really be very consistent with a Poisson.