Keep a strict probability condition strict at the quantile boundary

$\mu=(1.5+3.5)/2=2.5$ kg because the normal distribution is symmetric.

$P(H<1.5)=0.33$ gives $P((H-2.5)/\sigma<(1.5-2.5)/\sigma)=P(Z<-1/\sigma)=0.33$. The source quantile is $-1/\sigma=-0.43991$, so σ≈2.27 kg. It retains $H\sim N(2.5,2.2732^2)$ for later working.

Under independent selected crabs, the required probability is $P(1.5<H_1<3.5)P(1.5<H_2<3.5)P(H_3>3.5)P(H_4>3.5)(4!/(2!2!))=(1-0.33-0.33)^2(0.33)^2(4!/(2!2!))=0.0755$ to 3 significant figures. The source calls these observations X₁,…,X₄; they are observations of H.

Let $F\sim N(3.2,1.3^2)$ and $A=(H_1+H_2+F_1+F_2+F_3)/5$. Then $E(A)=(2.5+2.5+3.2+3.2+3.2)/5=2.92$ and $\operatorname{Var}(A)=(1/5)^2(2.2732^2+2.2732^2+1.3^2+1.3^2+1.3^2)\approx0.61620$. Thus $A\sim N(2.92,0.61620)$, using the source working precision.

$P(A>a)>0.6$ means $1-P(A\le a)>0.6$, hence $P(A\le a)<0.4$. Therefore a must be strictly below the 40 th-percentile boundary q≈2.721 kg. The source changes<to≤ and reports “largest a=2.72” to 3 significant figures. Its reported 2.72 kg is feasible, but there is no largest real a under the printed strict condition: the admissible set is a<q, with supremum q. If a is restricted to a 0.01 kg grid, the largest feasible grid value is 2.72 kg.