We have the standard deviation for the population, which means that the z-distribution is used to solve this question.
Doing this, we get that:
The 95% confidence interval for the population mean, in minutes, is (23, 26.5).
Sample mean:
The sample mean is the sum of all observations divided by the number of observations. We have that:
Thus, the sample mean is:
[tex]\overline{x} = \frac{10\times15 + 18\times30 + 12\times25}{10 + 18 + 12} = 24.75[/tex]
Confidence interval:
We have to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a p-value of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96\frac{5.64}{\sqrt{40}} = 1.75[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 24.75 - 1.75 = 23 minutes
The upper end of the interval is the sample mean added to M. So it is 24.75 + 1.75 = 26.5 minutes
The 95% confidence interval for the population mean, in minutes, is (23, 26.5).
A similar question is given at https://brainly.com/question/22668299
