Heres the question:
How can I find a probability given just mean and variance? Or do I need to use a distribution?
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0....please see my note below about the continuity correction. – 2011-11-28
1 Answers
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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0I'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 $\Pr(S\ge7.5)$. – 2011-11-28