Question

Six students measure the acceleration (in meters per second ) of an object in free fall. The measured values are: 10.56, 9.52, 9.73, 9.80, 9.78, 10.91. 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.

Answer 1

The value of $d$ is 0.86

Explanation

The measured values are......

10.56,  9.52,  9.73,  9.80,  9.78,  10.91

As there are total 6 values, so the Mean(μ) will be: $\frac{10.56+9.52+9.73+9.80+9.78+10.91}{6}= \frac{60.3}{6}=10.05$

The absolute deviation of each measured value $x$ from the mean(μ) is:  $|x-\mu|$

Thus.....

$|10.56-10.05|= 0.51\\ |9.52-10.05|= 0.53\\ |9.73-10.05|= 0.32\\ |9.80-10.05|= 0.25\\ |9.78-10.05|= 0.27\\ |10.91-10.05|= 0.86$

So, the greatest absolute deviation from the mean is 0.86 here. That means, $d=0.86$