Respuesta :
Answer:
a) 5.81
b) 25
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 5
Population standard deviation, σ = 13 milligrams
a) standard deviation of Juan's mean result
[tex]\text{Standard error} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{13}{\sqrt{5}} = 5.81[/tex]
5.81 is the standard deviation of Juan's mean result.
b) sample size such that the standard deviation reduces to 2.6
[tex]\text{Standard error} = \dfrac{\sigma}{\sqrt{n}}\\\\\Rightarrow \dfrac{13}{\sqrt{n}} = 2.6\\\\\Rightarrow \sqrt{n} = \dfrac{13}{2.6} = 5\\\\\Rightarrow n = 25[/tex]
Thus, Juan has to repeat the experiment 25 times to reduce the standard deviation to 2.6
Using the Central Limit Theorem, we have that:
a) The standard deviation is 5.81.
b) Juan must repeat the measurement 25 times.
The Central Limit Theorem states that for a sample of size n, from a population of standard deviation [tex]\sigma[/tex], the standard deviation of the sampling distribution is given by:
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that for the population, [tex]\sigma = 13[/tex].
Item a:
5 measurements, thus [tex]n = 5[/tex].
The standard deviation is:
[tex]s = \frac{13}{\sqrt{5}} = 5.81[/tex]
The standard deviation is 5.81.
Item b:
This is n for which [tex]s = 2.6[/tex], thus:
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
[tex]2.6 = \frac{13}{\sqrt{n}}[/tex]
[tex]2.6\sqrt{n} = 13[/tex]
[tex]\sqrt{n} = \frac{13}{2.6}[/tex]
[tex](\sqrt{n})^2 = (\frac{13}{2.6})^2[/tex]
[tex]n = 25[/tex]
Juan must repeat the measurement 25 times.
A similar problem is given at https://brainly.com/question/14174983
