We know by the empirical rule of the distribution under the normal curve falls within 3 standar deviations of the mean. This is:
[tex]\mu\pm3\sigma[/tex]In this case, we have the following:
[tex]\begin{gathered} \mu=11.60 \\ \sigma=0.15 \end{gathered}[/tex]Doing the substitution we have:
[tex]\begin{gathered} \mu\pm3\sigma \\ \Rightarrow11.60\pm3(0.15) \\ \Rightarrow11.60\pm0.45 \\ \Rightarrow(11.60-0.45,11.60+0.45) \\ \Rightarrow(11.15,12.05) \end{gathered}[/tex]therefore 95% of the amount of water dispensed is between 11.15 ounces and 12.05 ounces