Step 1
The \(\displaystyle{100}{\left({1}-\alpha\right)}\%\) confidence interval for the population mean, \(\displaystyle\mu\) when the population standard deviations, \(\displaystyle\sigma\) is known the z-confidence interval as follows:
\(\displaystyle{\left(\overline{{{x}}}-{Z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}\)
In this case,\(\displaystyle\overline{{{x}}},\sigma,\mu,\) are the sample mean, population standard deviation, population mean and sample size of the population, respectively and \(\displaystyle{z}_{{\frac{\alpha}{{2}}}}\) is the critical value of the standard normal distribution, above which,\(\displaystyle{100}{\left(\frac{\alpha}{{2}}\right)}\%{P}{S}{K}{\quad\text{or}\quad}{P}{S}{K}\frac{\alpha}{{2}}\) proportion of the observations lie.
Step 2
The sample mean is, \(\displaystyle\overline{{{x}}}={2.6}\), the population standard deviation is \(\displaystyle\sigma={0.3}\) and the sample size is \(\displaystyle{n}={36}\).
If 95% of the observations must lie within an interval, the remaining 5% must lie outside the interval.
Further, due to symmetry, 2.5% will lie above the upper limit and the remaining 2.5% will lie below the lower limit of the interval.
Thus, the upper limit of the interval is such that, 97.5% lie below it.
As a result,\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.025}}}\approx{1.96}\) [Using Excel formula: =NORM.S.INV(0.975)].
Thus, the confidence interval is,
\(\displaystyle{\left(\overline{{{x}}}-{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}={\left({2.6}-{\left({1.96}\right)}{\left({\frac{{{0.3}}}{{\sqrt{{{36}}}}}}\right)},{2.6}+{\left({1.96}\right)}{\left({\frac{{{0.3}}}{{\sqrt{{{36}}}}}}\right)}\right)}\)

\(\displaystyle={\left({2.6}-{\left({1.96}\right)}{\left({0.05}\right)},{2.6}+{\left({1.96}\right)}{\left({0.05}\right)}\right)}\)

\(\displaystyle{\left({2.6}-{0.098},{2.6}+{0.098}\right)}\)

PSK(2.502, 2.698) Thus, the 95% confidence interval for the mean zinc concentration in the river is (2.502 grams per liter, 2.698 grams per liter).

\(\displaystyle={\left({2.6}-{\left({1.96}\right)}{\left({0.05}\right)},{2.6}+{\left({1.96}\right)}{\left({0.05}\right)}\right)}\)

\(\displaystyle{\left({2.6}-{0.098},{2.6}+{0.098}\right)}\)

PSK(2.502, 2.698) Thus, the 95% confidence interval for the mean zinc concentration in the river is (2.502 grams per liter, 2.698 grams per liter).