Square the coefficient of a two-packet total and retain both tails

For $F\sim N(450,\sigma^2)$, $P(F\le437)=0.1$ gives $P((F-450)/\sigma\le(437-450)/\sigma)=P(Z\le-13/\sigma)=0.1$. The source uses $-13/\sigma=-1.2816$, obtaining σ≈10.144 g and reporting 10.1 g to 3 significant figures.

Let $S\sim N(500,9.2^2)$ and use $F\sim N(450,10.144^2)$ at the source working precision. Put $A=S_1+S_2+S_3+S_4+S_5+S_6-2(F_1+F_2)$. The requested probability is $P(|A|\ge1250)$, not just a one-sided tail.

$E(A)=E(S_1+\cdots+S_6)-2E(F_1+F_2)=6E(S)-2(2)E(F)=6(500)-4(450)=1200$.

Independence gives $\operatorname{Var}(A)=\operatorname{Var}(S_1+\cdots+S_6)+4\operatorname{Var}(F_1+F_2)=6\operatorname{Var}(S)+4(2)\operatorname{Var}(F)=6(9.2^2)+8(10.144^2)\approx1331.045$. Thus $A\sim N(1200,1331.045)$ at the source working precision.

$P(|A|\ge1250)=1-P(-1250<A<1250)=0.085268\approx0.0853$ to 3 significant figures. Absolute-value bars in both source probability lines are retained from the rendered pages.

The mass of each packet, for both spaghetti and fettuccine, must be independent of every other packet’s mass. This is the source assumption allowing the individual variances to be added.