Answer: 98.33 - 101.67
Step-by-step explanation:
The given Sample mean n= 100
Standard deviation[tex]\sigma=8[/tex]
Mean IQ in the sample [tex]\bar{x}=125[/tex]
Standard error of mean = [tex]\frac{\sigma}{\sqrt{\bar{x}}}[/tex]
Standard error of mean = [tex]\frac{8}{\sqrt{125}}[/tex]
Thus, standard error =[tex]\frac{8}{11.180}[/tex]
Standard error of mean (SE)= 0.7155
We know that z - score for 98% confidence interval is 2.33 .
Now, 98% confidence interval will be
n-z(SE) and n+z(SE)
[tex]=100-(0.7155)(2.33)\ and\ 100+(0.7155)(2.33)[/tex]
[tex]=100-1.66726\ and\ 100+1.66726[/tex]
[tex]=98.33274\ and\ 101.667[/tex]
Hence, the 98% confidence interval for the students' mean IQ score will be 98.33 - 101.67
