Using the Central Limit Theorem and the Empirical Rule, it is found that values of the sample mean between 62.88 and 65.12 would be consistent with the population distribution being N(64, 2.5).
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].
- By the Empirical Rule, 95% of the measures are within 2 standard errors of the mean.
In this problem:
- The distribution is N(64, 2.5), hence [tex]\mu = 64, \sigma = 2.5[/tex]
- The samples have 20 students, hence [tex]n = 20, s = \frac{2.5}{\sqrt{20}} = 0.56[/tex]
Applying the Empirical Rule:
64 - 2(0.56) = 62.88
64 + 2(0.56) = 65.12
Values of the sample mean between 62.88 and 65.12 would be consistent with the population distribution being N(64, 2.5).
A similar problem is given at https://brainly.com/question/25800303
