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).

Answer:The standard deviation of the sample mean decreases from 1.43 to 0.5 as n increases from 6 to 49.Step-by-step explanation:The standard deviation of the sample mean is given by the following formula:[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.In this problem, we have that:[tex]\sigma = 3.5[/tex]How is the standard deviation of the sample mean changed when the sample size is increased from n=6 to n=49?n = 6[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.5}{\sqrt{6}} = 1.43[/tex]n = 49[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.5}{\sqrt{49}} = 0.5[/tex]The standard deviation of the sample mean decreases from 1.43 to 0.5 as n increases from 6 to 49.