Answer
$mean=\mu_{\overline{x}}=60.5 cm$
$standard \ deviation=\sigma_{\overline{x}}=1.1cm$
Work Step by Step
The size of the sample(36) is 30 or more, therefore by the Central Limit Theorem the distribution of the mean ages is basically a normal distribution. Hence, $mean=\mu_{\overline{x}}=\mu=60.5 cm$
$standard \ deviation=\sigma_{\overline{x}}=\frac{\sigma}{\sqrt n}=\frac{6.6cm}{\sqrt{36}}=\frac{6.6cm}{6}=1.1cm$
