Question

Five students in Denver each make five movies of a ball falling through the air. From these movies, each student extracts a value of the local gravitational acceleration. Here are their results for the average and standard deviation.
--Student 1: 9.8685±±0.0051 m/s22
--Student 2: 9.7059±±0.0458 m/s22
--Student 3: 9.7086±±0.0329 m/s22
--Student 4: 9.8029±±0.0477 m/s22
--Student 5: 9.7399±±0.0125 m/s22
The National Geodetic Survey, whose measurements are linked back to the National Institute of Standards and Technology, reports that the correct value is 9.79597716±±0.00000042 m/s22.

What is the average of the student data set?

m/s22 ( ± 0.02 m/s22)

What is the standard deviation of the student data set?

Answer 1

the average of the student data set is

as mean = $dfrac{x_1+x_2+x_3+x_4+x_5}{5}$

variance = square of standard deviation and it is obtained by squaring the number subtracted from mean and adding them all

so variance =

standard deviation is square root of variance = 0.0624  

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