Using the z-distribution, it is found that the margin of error is of 0.059.
We are given the standard deviation for the population, hence the z-distribution is used.
What is the margin of error for a z-confidence interval?
It is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- [tex]\sigma[/tex] is the population standard deviation.
In this problem, the parameters are:
- [tex]\sigma = 0.25, n = 49[/tex].
- Confidence level of 0.90, hence, using a z-distribution calculator, the critical value is z = 1.645.
Then:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.645\frac{0.25}{\sqrt{49}}[/tex]
[tex]M = 0.059[/tex]
The margin of error is of 0.059.
You can learn more about the z-distribution at https://brainly.com/question/12517818