Given the sample mean is [tex] \mu=88 [/tex] and population standard deviation is [tex] \sigma=20 [/tex].
Now the standard deviation of the sample mean is [tex] \sigma_X=\frac{\sigma}{\sqrt{n}} =\frac{20}{\sqrt{36}}=3.3333 [/tex]
The [tex] (1-\alpha )100\% [/tex] confidence interval for population mean is
[tex] \mu_X \pm z_{\alpha /2}\sigma_X [/tex].
Here [tex] \alpha =0.1,z_{\alpha /2}=-1.645 [/tex]
The 90% CI for population mean is
[tex] 88 \pm 1.645 (3.3333)\\ (82.517,93.483) [/tex]
