Question

you measure the period of a mass oscillating on a vertical spring ten times as follows: period (s): 0.96, 1.40, 1.57, 1.25, 1.16, 1.46, 1.44, 1.50, 1.11, 1.22 what are the mean and (sample) standard deviation?

Answer 1

For the given sample, mean = 1.307 and standard deviation = 0.1865.

Given,

Sample

xi : 0.96, 1.40, 1.57, 1.25, 1.16, 1.46, 1.44, 1.50, 1.11, 1.22

Total samples N=10

Mean,

$x =$ (∑$x_{i}$) / N

= (0.96 + 1.40 + 1.57 + 1.25 + 1.16 + 1.46 + 1.44 + 1.50 + 1.11 + 1.22) / 10

 = 13.07/10

 = 1.307

Standard deviation =$\sqrt{}$((∑$x_{i}$ $-x)^{2}$/N)

Therefore we have to calculate (∑$x_{i} -x)^{2}$ first.

That is,

(∑$x_{i} -x)^{2}$

= (0.96 - 1.307)² + (1.40 - 1.307)² + (1.57 - 1.307)² + (1.25 - 1.307)² + (1.16 - 1.307)² + (1.46 - 1.307)² + (1.44 - 1.307)² + (1.50 - 1.307)² + (1.11 - 1.307)² + (1.22 - 1.307)²

= 0.34781

Now standard deviation = √((∑$x_{i} -x)^{2}$/N)

= $\sqrt{\frac{0.34781}{10} }$

= 0.1865

Therefore, for the given sample, mean = 1.307 and standard deviation = 0.1865.

Learn more about sample here: brainly.com/question/28443861

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