A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. How is the standard deviation of the sample mean changed when the sample size is increased from n=6 to n=49? Round all intermediate calculations to four decimal places (e.g. 12.3456) and round the final answer to three decimal places (e.g. 98.768).

Respuesta :

Answer:

n = 6: [tex]s = 1.4289[/tex]

n = 49: [tex]s = 0.5[/tex]

As the sample size is increased, standard deviation of the sample mean decreases.

Step-by-step explanation:

The standard deviation of the sample mean is the standard deviation off the population[tex]\sigma[/tex] divided by the square root of the size of the sample, that is [tex]n[/tex]. Mathematically, that is

[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with a standard deviation of 3.5 psi. So [tex]\sigma = 3.5[/tex]

So:

n = 6

[tex]s = \frac{3.5}{\sqrt{6}} = 1.4289[/tex]

n = 49

[tex]s = \frac{3.5}{\sqrt{49}} = 0.5[/tex]

As the sample size is increased, standard deviation of the sample mean decreases.