How can I find a probability given just mean and variance? Or do I need to use a distribution?

Heres the question:

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You've got a lot more than just the mean and variance: You've got 30 independent trials, so the central limit theorem can be used. –  Michael Hardy Nov 27 '11 at 16:32
....please see my note below about the continuity correction. –  Michael Hardy Nov 28 '11 at 0:30

This is a typical application of the Central Limit Theorem (or at least one form of it):

$S$ is a sum of independent, identically distributed variables. If you find the mean, $\mu$, and the standard deviation, $\sigma$, of $X_1$, then by the Central Limit Theorem, $S$ will have an approximately normal distribution with mean $30\mu$ and variance $30\sigma^2$.

The mean of $X_1$ is $$\mu=1\cdot {4\over 6}\cdot {1\over4} + {-1\over4}\cdot{4\over6}\cdot{3\over4}+0\cdot{2\over6}={1\over6}-{1\over8}={1\over24}.$$

Computing $\sigma^2$ for $X_1$:

$$\sigma^2=(1-{1\over24})^2\cdot {4\over 6}\cdot {1\over4} + {({-1\over4}-{1\over24})^2}\cdot{4\over6}\cdot{3\over4}+({1\over24})^2\cdot{2\over6} \approx 0.19618.$$

Then $$\mu_S=30\mu=30/24$$ and $$\sigma^2_S=30\sigma^2\approx 5.9375.$$

We can approximate the probability now: $$P[S\ge 8]\approx P\Biggl[Z\ge{ 8-{30\over24}\over \sqrt{ 5.9375} }\Biggr]\approx P[Z\ge{2.7701}]\approx0.002802.$$

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I am not sure that the variance calculation is correct. Isn't $$E[X_1^2] = 1^2\times\frac{1}{6} + \left(-\frac{1}{4}\right)^2\frac{3}{6} = \frac{19}{96}$$ so that the variance $E[X_1^2] - (E[X_1])^2$ is necessarily much smaller than $1$? –  Dilip Sarwate Nov 27 '11 at 16:40
@Dilip Sarwate Yes, thank you. Editing... –  David Mitra Nov 27 '11 at 16:44
In principle, we need quite a bit more than mean and variance. If we use the normal approximation, the estimated probability is small, and the true answer is small, so the difference between estimate and truth will be small. In that sense the approximation will be fairly good. If, however, we use ratio between truth and estimate as a criterion of accuracy, the performance of the normal approximation may not be good. –  André Nicolas Nov 27 '11 at 17:16
I'd use a continuity correction: The event $S\ge8$ is the same as $S>7$. When approximating an integer-valued discrete distribution with a continuous distribution, one conventionally uses the point half-way between those, and seeks $\Pr(S\ge7.5)$. –  Michael Hardy Nov 28 '11 at 0:29