In a series of measurements, the values of the acceleration due to gravity were 9.8 m/s2,9.7 m/s2,10.0 m/s2,9.9 m/s29.8\text{ m/s}^2, 9.7\text{ m/s}^2, 10.0\text{ m/s}^2, 9.9\text{ m/s}^2 and 9.6 m/s29.6\text{ m/s}^2. The mean absolute error is:

Model Answer & Options

Source: Extra Practice

0.12 m/s²

0.10 m/s²

0.50 m/s²

0.20 m/s²

Explanation

  1. Find the mean: gmean=(9.8+9.7+10.0+9.9+9.6)/5=49.0/5=9.8 m/s2g_{mean} = (9.8 + 9.7 + 10.0 + 9.9 + 9.6)/5 = 49.0 / 5 = 9.8\text{ m/s}^2. 2. Find absolute errors for each: 9.89.8=0,9.79.8=0.1,10.09.8=0.2,9.99.8=0.1,9.69.8=0.2|9.8-9.8|=0, |9.7-9.8|=0.1, |10.0-9.8|=0.2, |9.9-9.8|=0.1, |9.6-9.8|=0.2. 3. Mean absolute error = (0+0.1+0.2+0.1+0.2)/5=0.6/5=0.12 m/s2(0 + 0.1 + 0.2 + 0.1 + 0.2) / 5 = 0.6 / 5 = 0.12\text{ m/s}^2.

Try a Related Quiz

Test your skills on a similar concept: Units and Measurement - NCERT Class 11 Practice Set 1.

Start Related Quiz

Explore the Full Topic

This is just one question from the topic "error analysis".

View All Questions

Related Questions