Let $Y~N(\mu, 1)$. Use the fact that $P(\left | Y-\mu \right | < 1.96\sigma) \approx.95$ to construct an interval $(a(Y), b(Y))$, such that the probability $\mu$ is in the interval is approximately 0.95.
I am not sure where to start on this one. Any and all help is appreciated.
We will transform $|Y−μ|<1.96\sigma$ into the form $a(Y)<\mu<b(Y)$: $$|Y−μ|<1.96\sigma\Leftrightarrow1.96\sigma<Y−μ<1.96\sigma\Leftrightarrow1.96\sigma-Y<−μ<1.96\sigma-Y$$$$\Leftrightarrow Y-1.96\sigma<μ<Y+1.96\sigma.$$ From $P(\left | Y-\mu \right | < 1.96\sigma) \approx0.95$ and transforming done above we obtain $P(Y-1.96\sigma<μ<Y+1.96\sigma)\approx0.95.$