For a population with parameters as shown in the image below
(a) Mean of the distribution
[tex]\begin{gathered} \mu_{\bar{x}}\text{ = }\mu\text{ = 245.9} \\ \operatorname{mean}\text{ of the distribution = 245.9} \end{gathered}[/tex](b)Standard deviation of the distribution
[tex]\begin{gathered} \sigma_{\bar{x}\text{ }}\text{ = }\frac{\sigma}{\sqrt[]{n}} \\ \sigma_{\bar{x}\text{ }}\text{ = }\frac{16.7}{\sqrt[]{83}} \\ \sigma_{\bar{x}\text{ }}\text{ = }\frac{16.7}{9.11} \\ \sigma_{\bar{x}\text{ }}\text{ = 1.833} \\ \sigma_{\bar{x}\text{ }}\text{ = 1.83 (2 d.p)} \end{gathered}[/tex]Hence the value of the mean of the distribution = 245.9 and the standard deviation of the distribution = 1.83
