Respuesta :
when xbar is 48.5
-1 is (48.5 - 50)/(9/sqrt n)
-9/sqrt n is -1.5
n is (9/1.5²) is 36
when xbar is 51.5
+1 is (51.5 - 50)/(9/sqrt n)
9/sqrt n is1.5
n is (9/1.5)² is 36
-1 is (48.5 - 50)/(9/sqrt n)
-9/sqrt n is -1.5
n is (9/1.5²) is 36
when xbar is 51.5
+1 is (51.5 - 50)/(9/sqrt n)
9/sqrt n is1.5
n is (9/1.5)² is 36
Using the Normal distribution, the Empirical rule and the Central limit theorem, it is found that n is 36.
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.
- The empirical rule states that 68% of the measures are within 1 standard deviation of the mean.
- The central limit theorem states that for sampling distribution of sample means of size n, the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, since 68% of the students had sample means between 48.5 and 51.5 mm, we have that:
- When X = 51.5, Z = 1.
- When X = 48.5, Z = -1.
Using one of them, we can find the standard error. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1 = \frac{51.5 - 50}{s}[/tex]
[tex]s = 1.5[/tex]
Then, since [tex]\sigma = 9[/tex], the sample size is found solving the following equation:
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
[tex]1.5 = \frac{9}{\sqrt{n}}[/tex]
[tex]1.5\sqrt{n} = 9[/tex]
[tex]\sqrt{n} = \frac{9}{1.5}[/tex]
[tex](\sqrt{n})^2 = (\frac{9}{1.5})^2[/tex]
[tex]n = 36[/tex]
n is 36.
A similar problem is given at https://brainly.com/question/13002303
