Probability that the sum of 50 dice throws is $\geq 195$ The question is:
A cyclist is going for a bike journey of $1950$ kilometres. Every morning he tosses a die, and if the outcome is $k$, where $k = 1, 2,\dots,6$, he cycles $10k$ kilometres on that day. Find the probability that he will finish his route within $50$ days or less.
I took the question to mean what I've put in title (I don't know if that's correct), and when I attempted it I got about $0.889$ (edit I actually got $0.4562$, realised I made a mistake)
How do you answer this question?
 A: Your interpretation of the problem seems correct.
The mean of a single die throw is $\frac72$. The variance of a single die throw is $\frac{35}{12}$.
For $50$ die throws, these are multiplied by $50$: $175$ and $\frac{875}{6}$.
To approximate the Binomial Distribution $\ge195$ we compute the Normal Distribution $\frac{194.5-175}{\sqrt{\frac{875}{6}}}$ standard deviations above the mean.
The probability of being at least $1.61475$ standard deviations above the mean is approximately $5.318\%$.
A: You can work out this probability exactly with the following method.
The probability that the sum of $n$ dice is equal to $s$ is the coefficient on $x^s$ in the polynomial 
$$
\left(\frac{1}{6}x+\frac{1}{6}x^2+\frac{1}{6}x^3+\frac{1}{6}x^4+\frac{1}{6}x^5+\frac{1}{6}x^6\right)^n.$$
The probability that the sum of 50 dice is at least 195 is then the sum 
$$ \sum_{i=195}^{300} a_i$$
where $$ \left(\frac{1}{6}x+\frac{1}{6}x^2+\frac{1}{6}x^3+\frac{1}{6}x^4+\frac{1}{6}x^5+\frac{1}{6}x^6\right)^{50} = \sum_{i=50}^{300} a_i x^i.$$
With a computer algebra system, you can calculate that the probability is exactly 
$$ \frac{14335043218171457162351380541553904985}{269427092488254686881046533485512097792} = 0.053205648644... $$
A: Hint:  The expected value of one die is $3.5$, so for $50$ it is $175$.   The distribution is symmetric, so the chance must be less than $\frac 12$  What is the variance of one die?  What is the variance of $50$?  How many standard deviations do you need to be above the mean?
A: Let $X$ be the random variable that takes on the values of a six-sided die.
The question here about the probability the sum of $50$ throws exceeding $195$ is the same as asking the probability that the sample mean $\bar{X}$ exceeds $195/50=3.9$. 
The Central Limit Theorem says that the distribution of $\bar{X}$ is basically normal. So, compute $P(\bar{X}\geq 3.9)$ for the (practically) normally distributed variable $\bar{X}$. (The parameters of the distribution are derived from those of $X$, of course.)
