Answer:
[tex]1463.3\:<\:\mu\:<\:1580.7[/tex]
Step-by-step explanation:
Since the population standard deviation [tex]\sigma[/tex], is known, we use the z confidence interval for the mean.
This is given by:
[tex]\bar X-z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )\:<\:\mu\:<\bar X+z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]
For a 95% confidence interval we use [tex]z=1.96[/tex].
It was also given that: [tex]n=125[/tex], [tex]\bar X=1522[/tex] and [tex]\sigma=335[/tex]
Let us substitute the values to get:
[tex]1522-1.96(\frac{335}{\sqrt{125} } )\:<\:\mu\:<\:1522+1.96(\frac{335}{\sqrt{125} } )[/tex]
[tex]1463.3\:<\:\mu\:<\:1580.7[/tex]
Interpretation:
We can say with 95% confidence that the interval between 1463.3 and 1580.7 SAT scores contains the population mean based on the sample 125 SAT scores.
