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The question goes like this:

In an experiment, the time period of an oscillating object in five successive measurements is found to be $0.52$s, $0.56$s, $0.57$s, $0.54$s, $0.59$s. The least count of the watch used for the measurement of time period is $0.01$s. What is the percentage error in measurement of time period $T$.

My attempt: The maximum error in measurement of T due to limited precision of the measuring instrument is the least count i.e. $0.01$s. Also the mean of measured values is $$\frac { 0.52+0.56+0.57+0.54+0.59 }{ 5 } =0.556$$which when rounded off to 2 significant figures is $0.56$. Also the standard deviation can be calculated after rounding off as $0.02$, which can be a good estimate to random error. Hence, the value of $T$ can be written as $0.56\pm (0.01+0.02)=0.56\pm 0.03$s. Hence the percentage error should be $$\frac { 0.03 }{ 0.56 } \times 100\approx 5.357$$ Hence the percentage error should be $5.357$%.

But the answer given in the book is $3.57$%. How is this possible? Where did I commit a mistake?

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The least count of the watch used for the measurement of time period is $0.01$ s

This information is just telling you to round off to the second decimal place, as you correctly did.

The sample mean is $\mu = 0.56$ and the sample standard deviation is $\sigma = 0.02$. The answer the text is referring to is

$$\frac \sigma \mu = 0.0357 = 3.57 \%$$

But I would say that this is not entirely correct. The standard error is not $\sigma$, but

$$\frac \sigma {\sqrt N}$$

Where $N$ is the number of measurements. In our case,

$$\frac \sigma {\sqrt N}=0.009$$

So the real percentage error should be

$$\frac{0.009}{0.56} = 0.0161 = 1.61 \%$$

Update: a more careful discussion

As requested, I will try to explain more why we don't need to explicitly include the resolution of the instrument ($0.01/2$) in our calculation.

In my previous discussion I explained why the solution reported in your text was $35.7 \%$, but actually that reasoning is not really correct.

The sample mean of your data set is not really $\mu=0.56$, but $\mu=0.0556$, as you correctly wrote. But since they (incorrectly) used the standard deviation, $0.02$, as standard error, we have to round off the mean and write our result as

$$0.56 \pm 0.02$$

Because it would clearly be silly to write

$$0.556 \pm 0.02$$

because if we are not sure of the second decimal place we bother writing the third?

But if the correct standard error is used, we get

$$0.556 \pm 0.009$$

You may notice a strange thing: the number of significative digits has increased, even if our instrument had a resolution of only $0.01/2=0.005$. This is a property of the mean and it is why we use the mean in the first place: via the mean operation, we can increase the number of significative digits and circumvent the limitations of our instrument.

Take for example the case in which we have two measurements: $2$ and $7$, with resolution of $0.5$ clearly. The mean is $9/2=4.5$, so we have gained one significative place.

You can then see that with an infinite number of measurement our result becomes exact, regardless of the resolution of the instrument, because of the $\sqrt N$ term in the denominator of the standard error.

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  • $\begingroup$ I doubt this! Suppose instead of multiple readings, I just take one. Then should I say that there is no error? Surely, the precision of instrument not just tells about which digits to round off, but also the error. And of course, the idea to round off to two significant figures is evident from the given measured values themselves. $\endgroup$ – user104014 Jun 5 '16 at 8:50
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    $\begingroup$ The case where you have just one reading is different. In that case the only possible thing you can say is that the error is half the value of the last digit (in our case, it would be $0.005$), because for example $0.56$ could be $0.55511$ or $0.55999$ if measured with a better instrument. $\endgroup$ – valerio Jun 5 '16 at 8:58
  • $\begingroup$ Yes, but in case of multiple readings also, each reading carries an error as you described. So should we ignore it? $\endgroup$ – user104014 Jun 5 '16 at 9:20
  • $\begingroup$ That is not really an error: it is not due to fluctuations in the measuring process, it is a limit of your instrument. It would be better to call it the resolution of your instrument. The effect of the resolution is anyway already incorporated in the standard error, so you don't need to account for it separately. This is evident from the fact that the standard error goes to $0$ when $N$ tends to infinity, while the resolution is always the same. $\endgroup$ – valerio Jun 5 '16 at 12:03
  • $\begingroup$ I don't get it! How does the effect of resolution gets incorporated into the standard error? And if $N$ tends to infinity, then would you conclude that the mean of $N$ readings is exactly the true value 'regardless' of the resolution of instrument? $\endgroup$ – user104014 Jun 5 '16 at 14:34
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I think you are confusing systematic and random errors.
Your experimental results can give you no idea about the systematic error.
For example it might be that your timing device is calibrated incorrectly and when the correct time is 1.00 seconds then your timing device gives a reading of 1.10 seconds; when the correct time is 2.00 seconds the timing device gives a reading of 2.20 seconds.
Repeating readings or the smallest subdivision of your scale will not give you an indication of what the systematic error is.
You could only find that error by checking the calibration of your timing device against a reliable standard.

So in this example you have found an estimate of the random error by evaluation the standard deviation and that is the best you can do.

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  • $\begingroup$ Yes! My terminology was wrong. Edited. But even now, how do I INCORPORATE the error due to limited precision of the instrument. Surely, if I used a more precise instrument, my estimate of random error should be smaller. But how do I manipulate this mathematically? $\endgroup$ – user104014 Jun 5 '16 at 7:45
  • $\begingroup$ Just use the standard deviation. If you used an instrument with a finer scale division you might expect the standard deviation to be lower? $\endgroup$ – Farcher Jun 5 '16 at 8:08
  • $\begingroup$ Definitely not the standard deviation to be lower, but I doubt the statement that a finer scale division will not play any role in the error crept in our experiment. By the way is my answer right? $\endgroup$ – user104014 Jun 5 '16 at 8:23

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