SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Question : 184 of 200

Marks: +1, -0

An instrument is used to measure a quantity at different time instances, and the expected values of the quantity are to be the same; however, the measured values are different. A set of 5 measurements that were recorded at different time instances are 98, 101, 99, 100 and 102. Find the precision of the third measurement.

0.95
0.1
0.99
0.9

Validate

Solution:

The average measurement value

\ov { \X _{ \n }}={98+101+99+100+102}/{5}

={500}/{5}=100

Given third measurement reading = 99
The 3rd reading precision is given as

=1-|{99-100}/{100}|

=1-{1}/{100}=0.99

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