A toymaker makes $Xn$ specialized toy parts on day n, where $Xn$ are independent and identically distributed random variables with mean 6 and variance 9. Let $Sn$ be the total number of specialized parts produced from day 1 to day n. Using central limit theorem, determine the total number of parts, a, the said factory can guarantee to produce by day 50 with at least 99.9% certainty, i.e. determine the maximum value of a so that $P(S_{50}≥a)≥0.999$. The maximum value must be a whole number.
So I used the formula
$Zn = \frac{\sigma - \mu}{\sigma \sqrt n} $
and from z-table, $Zn = -3.10$
where,
$\sigma = \frac{Zn \sigma}{\sqrt n} + n \mu = \frac{(-3.10)(3)}{\sqrt 50} + (50)(6) = 298.69 \approx 299$
However my answer was incorrect. Any help would be appreciated.
In order to apply the central limit theorem you need to consider
$$\bar X_{50} = \frac 1{50}S_{50}$$
Hence, the probability in question is
$$P\left(\bar X_{50}\geq\frac a{50}\right) \geq 0.999$$
The standardization gives
$$Z \approx \frac{\bar X_{50} - \mu}{\sigma / \sqrt{50}}$$
Hence, you look for
$$P\left(Z \geq \frac{\frac a{50} - 6}{3 / \sqrt{50}}\right)\geq 0.999 \text{ or } P\left(Z < \frac{\frac a{50} - 6}{3 / \sqrt{50}}\right)\leq 0.001$$
Your $z$-score is correct: $z=-3.1$. So, you only need to find the highest integer $a$ with
$$\frac{\frac a{50} - 6}{3 / \sqrt{50}} < -3.1 \Leftrightarrow a < 234.239\ldots $$$$\Rightarrow \boxed{a=234}$$