You take $n = 50$ dice rolls and calculate the average value of these dice rolls. What is the probability (approximately -i.e using CLT) that this average is greater than or equal to $4$?
My attempt:
We know that each die has expectation and $3.5$ and variance $35/12$. So the sum has variance $3.5n = 105$ and variance $\frac{35}{12}n = 145.8\overline{3}$. So, I was thinking to evaluate the cumulative probability of a normal distribution with parameters $\mu = 105, \sigma^2 = 145.8\overline{3}$, however I'm not sure if this is correct.
Best Answer
For completeness, your normal approximation might suggest a probability of about $$1-\Phi\left(\dfrac{4-3.5}{\sqrt{(35/12)/50}}\right) \approx 0.019217$$ where $\Phi$ is the cumulative distribution function of a standard normal distribution
This is not bad as the actual probability is about $0.021086$, but you can do slightly better with a continuity correction. A mean of at least $4$ corresponds to a sum of at least $200$ or a sum strictly more than $199$ on this discrete distribution, so checking $4$ or $\frac{199}{50}=3.98$ are both arguable on the continuous normal distribution, and a compromise would be to check $3.99$ so about the closer $1-\Phi\left(\tfrac{4-3.5}{\sqrt{(35/12)/50}}\right) \approx 0.021240$
As a warning, a normal approximation like this can be relatively poor in the tails of a distribution. If for example you had been looking for the probability of a mean of at least $5$, the actual probability would have been about $5.8 \times 10^{-11}$ but a straight normal approximation might have been about $2.6\times 10^{-10}$ and with continuity correction about $3.4\times 10^{-10}$, both several times too high