Answer:
[0.9 months, 32.69 months]
Step-by-step explanation:
The mean is
[tex]\bf \bar x=\frac{5+17+12+21+29}{5}=16.8[/tex]
The standard deviation is
[tex]\bf s=\sqrt{\frac{(5-16.8)^2+(17-16.8)^2+(12-16.8)^2+(21-16.8)^2+(29-16.8)^2}{5}}=8.1092[/tex]
Now, we have to find two values a and b such that the area under the Normal curve with mean 16.8 and standard deviation 8.1092 between a and b equals 95% = 0.95
Using a spreadsheet we find these values are
a = 0.906
b = 32.694
and our 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program rounded to two decimal places is
[0.9 months, 32.69 months]