Respuesta :
Using the normal distribution and the central limit theorem, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).
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].
In this problem:
- The mean is of [tex]\mu = 3.5[/tex].
- The standard deviation is of [tex]\sigma = 0.5[/tex].
- Sample of 100, hence [tex]n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05[/tex]
The interval that contains 95.44% of the sample means for male students is between Z = -2 and Z = 2, as the subtraction of their p-values is 0.9544, hence:
Z = -2:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-2 = \frac{X - 3.5}{0.05}[/tex]
[tex]X - 3.5 = -0.1[/tex]
[tex]X = 3.4[/tex]
Z = 2:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]2 = \frac{X - 3.5}{0.05}[/tex]
[tex]X - 3.5 = 0.1[/tex]
[tex]X = 3.6[/tex]
The interval that contains 99.44% of the sample means for male students is (3.4, 3.6).
You can learn more about the normal distribution and the central limit theorem at https://brainly.com/question/24663213
