Given:
mean, μ = 11 lb
Std. deviation, σ = 2.1
Let x = the weight that separates the top 5% of the crop.
The z-score is
z = (x - μ)/σ
= (x - 11)/2.1
From standard z-table for normal distribution,
P(z=1.645) = 0.95
Therefore
(x - 11)/2.1 = 1.645
x - 11 = 1.645/2.1 = 0.7833
x = 11.783
Answer: 11.8 lb (nearest tenth)