P value for a z-score of 4.9? Or am I doing this wrong? My question is as follows:
A fair die is rolled $120$ times. Find the probability that $5$ is on the 
top:
a. between $30$ and $40$ times,
b. between $18$ and $50$ times,
c. more than $70$ times.
Hint: Use the approximation of the Binomial distribution to the normal distribution.
I am close to solving it. However, I am stuck, since I obtained a standard deviation value of $4.0825$ and therefore one of the $z$-scores as $4.9$. My solution is as follows:
$x$ = event of having a $5$ shown on top
$P(x) = 1/6$
This is a binomial probability experiment
$np(1-p) \ge 10$?
$120 * \frac{1}{6} ( 1 - \frac{1}{6}) = \frac{50}{3} > 10$
mean $= np = 120$; 
standard deviation $= \sqrt{\frac{50}{3}} = 4.0825.$
$X~N(\mu=20, \sigma=4.0825)$
a) $P(30\le x\le 40) =$ ?
$z = \frac{x - \mu} {\sigma} $
$P\left(\frac{30 - 20}{4.0825} \le z \le \frac{40 - 20}{4.0825} \right)= P (2.45\le z\le4.9)$
At this point, I am stuck, since the $p$-value for a $Z$-score of $4.9$ is impossible to find (an online calculator gave me a value of $0.9999995$). Is my solution wrong or should I proceed with the given $p$-value?
Thank you all. 
: )
 A: I wanted to write this in a comment but got carried away and ran out of characters:
Your working all looks pretty fine to me (although you have missed out the end of the line where you work out the mean, but you then do use the correct value ($20$) in your workings). Also I think it's important that typically we write the Normal distribution as $X \sim N(\mu, \sigma^2)$, not $X\sim N(\mu, \sigma)$. Again, you've used the correct values in your work though.
I'm guessing that the low probability is what is making this difficult. Typically to approximate the Binomial to the Normal you want $p \approx \frac 12$. Whereas here obviously we have a much lower $p$. 
I think in your shoes I would take the value to be $\approx 1$, and write a small note alongside my answer explaining why I've done this. If this is a homework question then I'm sure your teacher would be more than happy to explain what there thinking is about this.
A: As a check of your work, the exact probabilities are given as follows:
> sum(dbinom(30:40,120,1/6))
[1] 0.01287915
> sum(dbinom(18:50,120,1/6))
[1] 0.7240251
> sum(dbinom(71:120,120,1/6))
[1] 1.112216e-25

You will notice that these probabilities are slightly higher than the ones you obtained since you are not using a continuity correction. To be more precise, you could take
$$P(30\leq X\leq 40)\approx \Phi\left(\frac{40.5-20}{4.0825}\right)-\Phi\left(\frac{29.5-20}{4.0825}\right)=0.00998$$
> pnorm(20.5/4.0825)-pnorm(9.5/4.0825)
[1] 0.00998223

and similarly for the other problems.
