Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 49 - 1 = 48 df, is t = 2.0106.
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 12.2 - 2.0106\frac{1.6}{\sqrt{49}} = 11.7[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 12.2 + 2.0106\frac{1.6}{\sqrt{49}} = 12.7[/tex]
The 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).
More can be learned about the t-distribution at https://brainly.com/question/16162795