Solution:

Given information:

\(\displaystyle{n}={125}\) Sample size

\(\displaystyle{x}={64.5}\) inches Sample mean

\(\displaystyle{s}={5}\) inches sample standard deviation

\(\displaystyle\alpha={0.05}\) Level of significance

Step 2

The 95% confidence interval for the mean is

\(\displaystyle\overline{{{x}}}\pm{t}_{{\frac{\alpha}{{2}},{n}-{1}}}\times{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

n is large we used \(\displaystyle{z}_{{\frac{\alpha}{{2}}}}\) instead of \(\displaystyle{t}_{{\frac{\alpha}{{2}},{n}-{1}}}\)

At \(\displaystyle\alpha={0.05}\)

\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.05}}}={1.96}\) From Z table

\(\displaystyle{\left({64.5}\pm{1.96}\times{\frac{{{5}}}{{\sqrt{{{125}}}}}}\right)}\)

\(\displaystyle{\left({64.5}\pm{1.96}\times{\frac{{{5}}}{{{11.180339}}}}\right)}\)

\(\displaystyle{\left({64.5}\pm{1.96}\times{0.4472135}\right)}\)

\(\displaystyle{\left({64.5}\pm{0.8765384}\right)}\)

\(\displaystyle{\left({64.5}-{0.8765384},{64.5}+{0.8765384}\right)}\)

\(\displaystyle{\left({63.62346},{65.376538}\right)}\)

\(\displaystyle{\left({63.62},{65.38}\right)}\)

The \(\displaystyle{95}\%\) confidence interval for the mean is ( 63.62, 65.38)