Given that

Sample size, \(\displaystyle{n}={51}\)

Sample mean \(\displaystyle\overline{{x}}={8.19}\)

Sample standard deviation, \(\displaystyle{s}={1.45}\)

The objective is to construct \(95\%\) large-sample confidence intervals for the true average elongation \(\displaystyle\mu\).

The confidence level is,

\(\displaystyle{C}={95}\%=\frac{95}{{100}}={0.95}\)

The level of significance is,

\(\displaystyle\alpha={1}-{C}={1}-{0.95}={0.05}\)

From standard normal table, at 0.05 level of significance the two-tailed critical value is,

\(\displaystyle{z}_{{\alpha\text{/}{2}}}={z}_{{{0.05}\text{/}{2}}}={1.96}\)

Step 2

The confidence interval is computed below.

\(\displaystyle\overline{{x}}\pm{z}_{{\alpha\text{/}{2}}}{\left(\frac{s}{\sqrt{{n}}}\right)}\)

\(\displaystyle{8.19}\pm{z}_{{{0.05}\text{/}{2}}}{\left(\frac{1.45}{\sqrt{{51}}}\right)}\)

\(\displaystyle{8.19}\pm{1.96}{\left({0.2030}\right)}\)

\(\displaystyle{8.19}\pm{0.39788}\)

\(\displaystyle{\left({8.19}-{0.39788},{8.19}+{0.39788}\right)}\)

\(\displaystyle{\left({7.79212},{8.58788}\right)}\)

\(\displaystyle{\left({7.792},{8.588}\right)}\) (rounded to three decimals)