Answer: (80.1786, 83.8214)
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}\pm z_c\dfrac{\sigma}{\sqrt{n}}[/tex] , where [tex]\overline{x}[/tex] = Sample mean , n= sample size , [tex]\sigma[/tex] = standard deviation, [tex]z_c[/tex] = critical z-value.
Given: n= 50, [tex]\overline{x}= 82[/tex] , [tex]\sigma=5[/tex]
Critical z-value for 99% confidence level = 2.576
Now, a 99 % confidence interval for the mean:
[tex]82\pm(2.576)\dfrac{5}{\sqrt{50}}\\\\=82\pm(2.576)(0.70710)\\\\=82\pm1.8215\\\\=(82-1.8214,\ 82+1.8214)\\\\=(80.1786,\ 83.8214)[/tex]
Hence, required 99% confidence interval = (80.1786, 83.8214)
