So far I have used Chebychev's inequality to calculate $n = 157$, and my initial thinking for applying the central limit theorem is $-1.2815 = (5-0n)/(n^\sqrt{391/n}) $because $P(Z > -1.2815) = 90%$, however this isn't giving me the correct value for $n$ so I'm not sure what to try now.
"You want to determine the melting point $c$ of a new material. You have $n$ specimens on each of which you make a measurement of the melting point in degrees Kelvin, giving you a dataset $m_1,…,m_n$. We model this with random variables $M_i=c+U_i$, where $U_i$ is the random measurement error. It is known that $E[U_i]=0$ and $Var(U_i)=391$ for each $i$, and that we may consider the random variables $M_1,M_2,…$ as independent.
how many measurements do you need to perform to be 90% sure that the average of the measurements is within 5 degrees of c?
Use the normal approximation (central limit theorem rule of thumb) to find a value for n . You can use that the 0.95 quantile of the standard normal distribution is q(0.95)≈1.645 ."
\begin{align} P \left(\left| \frac{\sum_{i=1}^nM_i}{n}-c\right|<5 \right)&=P \left(\sqrt{n}\left| \frac{\sum_{i=1}^nM_i}{n}-c\right|<5\sqrt{n} \right)\\ &\approx P\left(|Z|<\frac{5\sqrt{n}}{\sqrt{Var(U_i)}}\right) \end{align}
That is we want $$\frac{5\sqrt{n}}{\sqrt{Var(U_i)}}\ge 1.645$$