probability - chance of X throws of a die totalling Y Im studying probability in my own time on an MOOC. Problem with the MOOC is often the homework answers are not explained at all, which makes learning from them difficult.
I have one such problem i need help understanding.
I have 1 die. I throw it 50 times. What is the probability that the total of sum of each throw will be $\ge 230$?
I have tried playing around with binomial distribution & normal distribution statistics to figure it out but have not gotten my head around it yet, and would appreciate some help.
 A: There's not a very straightforward way to do this using just the binomial distribution. To see why, remember that the binomial distribution counts successes out of $n$ attempts at a Bernoulli experiment. Informally, you have to ask a yes/no question $n$ times, with the probability of a "yes" being the same on each ask, and you want to count the yes's. Adding up the spots on 50 rolls of a die is more complicated than that.
Now, the expected value of each roll is $3.5$. (Do you know why?) The variance of a single roll is $2.91\overline{6}$. (Do you know why?) You can mulitply those by $50$ to get the expected value and variance of $50$ rolls. That distribution should be approximately normal. (Do you know why?) Given its expected value and standard deviation, you should be able to answer your question.
Does this help?

By the way, to do this without a normal approximation, you would have to count up the number of ways to roll $\geq 230$ with $50$ rolls, and then multiply that count by $\left(\frac16\right)^{50}$. That combinatorics problem would be... arduous, but not impossible.
A: The expected value and variance of the result of throwing a die once are respectively $3.5$ and $35/12\approx 2.91666\ldots.$ Throw the die $50$ times and you multiply those by $50.$
Be careful about the following:
\begin{align}
\frac{35}{12} \times 50 & \approx 2.91667\times50 & & = 145.835 \\[10pt]
\frac{35}{12} \times 50 & = \frac{35} 6 \times 25 = \frac{875} 6 & & = 145.8333\ldots
\end{align}
Don't round until the last step except when you know how it will affect the bottom line. Notice that the last digit in the rounded answer is wrong: you get $5$ where you should get $3$.
Next let us treat the continuity correction: The normal distribution is a continuous random variable which we will use to approximate a discrete random variable. Notice that with the integer-valued sum the following say the same thing:\begin{align} \text{sum} & \ge 230 \\ \text{sum} & > 229 \end{align}
With the continuous random variable we will use the number half way between those $229.5.$
Now we have
\begin{align}
\Pr( \text{sum} \ge 229.5 ) & = \Pr\left( \frac{\text{sum} - (3.5\times50)}{\sqrt{875/6}} \ge \frac{229.5 - (3.5\times50)}{\sqrt{875/6}} \right) \\[10pt]
& = \Pr\Big( Z \ge {[}\text{insert number here}{]} \Big) \\[10pt]
& = {[}\text{get this number from the table or from your software}{]}
\end{align}
