Respuesta :
Using the z-distribution, it is found that she should take a sample of 97 scores to ensure that the margin of error is below 20.
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 population.
The margin of error is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
Scores on the math portion of the SAT are believed to be normally distributed and range from 200 to 800, hence by the Empirical Rule the standard deviation is given by:
[tex]6\sigma = 800 - 200[/tex]
[tex]\sigma = 100[/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.
To ensure a margin of error of 20, we solve for n when M = 20, hence:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]20 = 1.96\frac{100}{\sqrt{n}}[/tex]
[tex]20\sqrt{n} = 1.96 \times 100[/tex]
[tex]\sqrt{n} = 1.96 \times 5[/tex]
[tex](\sqrt{n})^2 = (1.96 \times 5)^2[/tex]
n = 96.04.
Rounding up, she should take a sample of 97 scores to ensure that the margin of error is below 20.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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