# A study a local high school tried to determine the mean height of females in the US. A study surveyed a random sample of 125 females and found a mean height of 64.5 inches with a standard deviation of 5 inches. Determine a 95\% confidence interval for the mean.

A study a local high school tried to determine the mean height of females in the US. A study surveyed a random sample of 125 females and found a mean height of 64.5 inches with a standard deviation of 5 inches. Determine a $$\displaystyle{95}\%$$ confidence interval for the mean.
$$\begin{array}{|c|c|} \hline \text{Confidence Interval} & z \\ \hline 80\% & 1.282 \\ \hline 85\% & 1.440\\ \hline 90\% & 1.645\\ \hline 95\% & 1.960\\ \hline 99\% & 2.576\\ \hline 99.5\% & 2.807\\ \hline 99.9\% & 3.291\\ \hline \end{array}$$

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Step 1
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)