A student performs an experiment to measure the diameter of a wire and obtains the following readings: 1.25 cm, 1.26 cm, 1.24 cm, 1.25 cm, 1.27 cm. The mean absolute error in the measurement is:

6 views 1 helpful Updated Jul 26, 2026
Appears in NEET UG
Solution ✔ Verified
  • A0.008 cm
  • B0.01 cm
  • C0.012 cm
  • D0.02 cm
Explanation

Mean reading = (1.25 + 1.26 + 1.24 + 1.25 + 1.27) / 5 = 6.27 / 5 = 1.254 cm. Rounding the mean to the same number of decimal places as the measurements (two decimal places) gives 1.25 cm. Individual absolute errors: |1.25 - 1.25|=0.00, |1.26 - 1.25|=0.01, |1.24 - 1.25|=0.01, |1.25 - 1.25|=0.00, |1.27 - 1.25|=0.02. Sum of absolute errors = 0.00 + 0.01 + 0.01 + 0.00 + 0.02 = 0.04 cm. Mean absolute error = 0.04 / 5 = 0.008 cm.

Was this solution helpful?
1
Practice MCQs on Errors in Measurement — test yourself with instant answers. Start Practising