Using the Central Limit Theorem, it is found that a sample size of 13 is needed.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For this problem, using a calculator:
- The mean is [tex]\mu = 37.9[/tex]
- The standard deviation is [tex]\sigma = 14.07[/tex]
The sample size needed is n when [tex]s = 4[/tex], as by the Empirical Rule, 95% of the measures are within 2 standard errors of the mean, hence [tex]2s = 8 \rightarrow s = 4[/tex]
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
[tex]4 = \frac{14.07}{\sqrt{n}}[/tex]
[tex]4\sqrt{n} = 14.07[/tex]
[tex]\sqrt{n} = \frac{14.07}{4}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{14.07}{4}\right)^2[/tex]
[tex]n = 12.37[/tex]
Rounding up, a sample size of 13 is needed.
A similar problem is given at https://brainly.com/question/24663213
