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The bore diameters of eight randomly selected bearings are shown here (in mm): 50.001, 50.002, 49.998, 50.006, 50.005, 49.996, 50.003, 50.004 a) Calculate the sample average. B) Calculate the sample standard deviation.

Respuesta :

Answer:

sample average = 50.00187

sample standard deviation = 0.0034

Explanation:

given data

8 diameters = 50.001, 50.002, 49.998, 50.006, 50.005, 49.996, 50.003, 50.004

solution

first we get her sample average that is express as

sample average = [tex]\frac{\sum x}{n}[/tex]      ...............1

here x is diameter of bearing

and n is total no of sample

put here value and we get

sample average = [tex]\frac{400.015}{8}[/tex]

sample average = 50.00187  

and

now we get here sample standard deviation

sample standard deviation = [tex]\sqrt{\frac{(x-\bar x )^2}{n- 1})[/tex]     ...............2

put here value and we get

sample standard deviation = [tex]\sqrt{\frac{00008288}{8- 1})[/tex]

sample standard deviation = 0.0034

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