Maximise a normal interval probability by choosing its mean

The variance is 16, so the standard deviation is 4. A normal density is symmetric and strictly decreases away from its mean. Among intervals of this fixed width, the probability is greatest when the endpoints are equally distant from the mean. Thus choose $\mu=(-3.74+5.82)/2=1.04$; the half-width is 4.78.

Independently, write $g(\mu)=\Phi((5.82-\mu)/4)-\Phi((-3.74-\mu)/4)$, where $\Phi$ is the standard-normal CDF and $\phi$ its density. Then $g^{\prime}(\mu)=[\phi((-3.74-\mu)/4)-\phi((5.82-\mu)/4)]/4$. This derivative is positive for $\mu<1.04$, zero at 1.04 and negative for $\mu>1.04$, proving the global maximum.

At $\mu=1.04$, standardising gives $P(-1.195\le Z\le1.195)=2\Phi(1.195)-1\approx0.7679129699$. The source short answer is 0.768, consistent with three-significant-figure rounding. The source prints no separate rounding requirement; the derivation and fuller decimal here are independently authored.