Construct the lower bound of a confidence interval using the following information: $n=18$; mean = 51.056; standard deviation $=32.288$; alpha $=0.05$. Remember that alpha $=0.05$ is for a $95\mathrm{\%}$ confidence interval and alpha $=0.01$ is for a $99\mathrm{\%}$ confidence interval.

Step 1 ：Given the following information: sample size $n=18$, mean $=51.056$, standard deviation $=32.288$, and alpha $=0.05$. Remember that alpha $=0.05$ is for a $95\mathrm{\%}$ confidence interval.

Step 2 ：To construct the lower bound of a confidence interval, we need to use the formula: mean - Z * (standard deviation / sqrt(n)), where Z is the Z-score corresponding to the desired confidence level.

Step 3 ：For a $95\mathrm{\%}$ confidence interval, the Z-score is approximately $1.96$.

Step 4 ：Substitute the given values into the formula to find the lower bound of the confidence interval: $51.056-1.96\ast (32.288/\sqrt{18})$.

Step 5 ：Calculate the above expression to get the lower bound of the confidence interval, which is approximately $36.14$.

Step 6 ：Final Answer: The lower bound of the $95\mathrm{\%}$ confidence interval is $\boxed{{\displaystyle 36.14}}$.