Sample size = n = 81
Population Standard Deviation = s = 45
Sample Mean = x = 112
Since, population standard deviation is known, we will use z distribution to find the confidence interval.
The critical z value for 99.7% is 2.968
So, the confidence interval will be:
[tex](112-2.968* \frac{45}{ \sqrt{81} }, 112-2.968* \frac{45}{ \sqrt{81} }) \\ \\
(97.16, 126.84)[/tex]
Rounding of to nearest integers the confidence interval will be 97 to 127.
So, option A gives the correct answer
