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Sagot :
Using the z-distribution, as we have the standard deviation for the population, the margin of error is given by:
b. 0.39.
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- z is the critical value.
- n is the sample size.
- [tex]\sigma[/tex] is the standard deviation for the sample.
The margin of error is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
As for the other parameters, we have that [tex]\sigma = 1.8, n = 81[/tex].
Hence, the margin of error is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.96\frac{1.8}{\sqrt{81}}[/tex]
[tex]M = 0.39[/tex]
Hence option b is correct.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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