Question

Six students measure the acceleration (in meters per second per second) of object in free fall. The measured values are shown. The students want to state that the absolute deviation of each measured value x from the mean is at most d. Find the value of d. 10.56, 9.52, 9.73, 9.80, 9.78, 10.91

Answer 1

The measured data is
x=[10.56, 9.52, 9.73, 9.80, 9.78, 10.91].

Calculate the mean.
m = (10.56+9.52+9.73+9.80+9.78+10.91)/6 = 10.05

Calculate the deviations from the mean.
k = x - m
   = [0.51, -0.53, -0.32, -0.25, -0.27, 0.86]

Calculate the absolute deviation from the mean.
$d= \frac{1}{6} | (0.51-0.53-0.32-0.25-0.27+0.86) | = |1.184\times10^{-15}| \approx 0$

Answer: 0