Error Analysis involving random errors 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?
 A: 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.
A: 
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.
