Infer a normal standard deviation from symmetric tail areas

By the CLT, $\bar C\mathrel{\dot\sim}N(2.5,0.4^2/n)$ for a large independent sample. Require $P(Z>-0.05\sqrt n/0.4)>0.95$, so $0.05\sqrt n/0.4>\Phi^{-1}(0.95)\approx1.64485$. Thus $n>173.1\ldots$ and the least integer is 174.

Let $a=P(X<2.5)$. The cutoffs 2.5 and 2.9 are equally spaced from the mean 2.7, so the right tail also has area $a$, while $P(X<2.9)=(107/50)a$. The middle area is $[(107/50)-1]a$. Hence $a+[(107/50)-1]a+a=1$, giving $a=50/157$. Thus $-0.2/\sigma=\Phi^{-1}(50/157)=-0.4719777448\ldots$, so $\sigma=0.4237487952\ldots\approx0.424$ kg.

Using the printed rounded $\sigma=0.424$, one duck’s cost is $Y=4X\sim N(10.8,4^2(0.424)^2)=N(10.8,2.876416)$. For independent ducks, $Y_1+Y_2\sim N(21.6,5.752832)$. Having less than $27.50 left means spending more than $ 22.50, so the probability is $P(Y_1+Y_2>22.50)\approx0.354$.

The maximum cost is below 11 exactly when every duck costs below 11. Independence gives $[P(Y<11)]^4=[P(X<2.75)]^4\approx0.0895$. Carrying the unrounded $\sigma$ gives about 0.0895024, which has the same printed 3 s.f. value.