Question

you measure the period of a mass oscillating on a vertical spring ten times as follows: period (s): 1.38, 1.18, 1.47, 1.35, 1.52, 1.23, 1.62, 1.17, 1.03, 1.11 what are the mean and (sample) standard deviation?

Answer 1

For the given sample, mean = 1.306 and standard deviation = 0.1826.

As given ,

Given sample

$x_{i}$: 1.38, 1.18, 1.47, 1.35, 1.52, 1.23, 1.62, 1.17, 1.03, 1.11

Total samples N=10

Mean

 $\bar{x} =\frac{\sum x_{i} }{N}$

    = ( 1.38+1.18+ 1.47+ 1.35+ 1.52+ 1.23+ 1.62+ 1.17+ 1.03+1.11)/ 10

    = (13.06 ) /10

    = 1.306

$(\sum x_{i} - \bar{x} ) ^{2} }$  

= (1.38 -1.306)² + (1.18 -1.306)² +(1.47 -1.306)² +(1.35 -1.306)² +(1.52 -1.306)² +(1.23 -1.306)² +(1.62 -1.306)² +(1.17-1.306)² +(1.03 -1.306)² +(1.11 -1.306)²

=0.33344

Standard deviation =$\sqrt{ \frac{(\sum x_{i} - \bar{x} ) ^{2} }{N} }$

                              = $\sqrt{\frac{0.33344}{10} }$

                              = 0.1826

Therefore, for the given sample, mean = 1.306 and standard deviation = 0.1826.

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