Answer: [tex](2.8,\ 3.6)[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
Given : Sample size : n= 51 ( >30 , that means its a large sample)
Sample mean : [tex]\overlien{x}=3.2\text{ grams}[/tex]
Standard deviation : [tex]\sigma=1.1\text{ grams}[/tex]
Significance level : [tex]1-0.99=0.01[/tex]
Critical value : [tex]z_{\alpha/2}=\pm2.576[/tex]
Now, the 99% confidence interval for the mean grams of sugar will be :-
[tex]=3.2\pm(2.576)\times\dfrac{1.1}{\sqrt{51}}\\\\\approx3.2\pm0.40\\\\=(2.8,\ 3.6)[/tex]
Hence, the 99% confidence interval for the mean grams of sugar [tex]\mu[/tex] = [tex](2.8,\ 3.6)[/tex]