Respuesta :
Using the z-distribution, a sample of 609 nickels has to be weighed.
What is a z-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
The margin of error is:
[tex]M = 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 population.
For this problem, the parameters are:
[tex]M = 20, \sigma = 300, z = 1.645[/tex]
We solve for n to find the sample size, then:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]20 = 1.645\frac{300}{\sqrt{n}}[/tex]
[tex]20\sqrt{n} = 1.645 \times 300[/tex]
[tex]\sqrt{n} = 1.645 \times 15[/tex]
[tex](\sqrt{n})^2 = (1.645 \times 15)^2[/tex]
n = 608.9
Rounding up, a sample of 609 nickels has to be weighed.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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