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Sagot :
Using the z-distribution, it is found that the margin of error for a 95% confidence interval is of $0.04.
What is a z-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, the values of the parameters are:
[tex]z = 1.96, \sigma = 0.48, n = 576[/tex].
Hence, the margin of error, in dollars, is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.96\frac{0.48}{\sqrt{576}}[/tex]
M = 0.04.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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