If we assume that each of the trials had a normally distributed standard error of 0.05 mL, and that all of these errors were independent of each other, then the correct way to combine the errors is via quadrature. The idea here is that the average $\bar{x}$ will depend on each of the measurements $x_1, x_2, \dots$ by $$ \bar{x} = \frac{1}{5} (x_1 + x_2 + x_3 + x_4 + x_5). $$ If each $x_i$ has an uncertainty $\Delta x_i$, then the general formula for the uncertainty in the derived quantity $\bar{x}$ will be $$ (\Delta \bar{x})^2 = \sum_i \left( \frac{\partial \bar{x}}{\partial x_i} \Delta x_i \right)^2. $$ In the present case, all of the $\Delta x_i$ values are the same, and this works out to be $$ \Delta \bar{x} = \sqrt{ 5 \cdot \frac{1}{25} (\Delta x)^2} = \frac{\Delta x}{\sqrt{5}} \approx 0.0223 \, \text{mL}. $$ Note that this implies that as as you take more and more repeated measurements, you should get a better and better idea of what the "real" value is, since $\Delta \bar{x}$ is decreasing with time.

That said, there are some caveats to this technique. In particular, it makes a pretty strong (if standard) assumption about your measurement technique: namely, that you're equally likely to get a value that's too high or too low, and that your errors are distributed in some sort of bell curve about some "true" value. The method described here still holds (I think) under certain relaxations of the latter assumption. The former assumption, however, is pretty necessary for this logic to work; and it's easy to think of a situation where it might not hold. (In particular, "last-digit" measurement error might tend to be a bit too high or too low depending on many factors.) Proper error analysis requires a lot of careful thought; there's no one-size-fits-all approach to it.

If we assume that each of the trials had a normally distributed standard error of 0.05 mL, and that all of these errors were independent of each other, then the correct way to combine the errors is via quadrature. The idea here is that the average $\bar{x}$ will depend on each of the measurements $x_1, x_2, \dots$ by $$ \bar{x} = \frac{1}{5} (x_1 + x_2 + x_3 + x_4 + x_5). $$ If each $x_i$ has an uncertainty $\Delta x_i$, then the general formula for the uncertainty in the derived quantity $\bar{x}$ will be $$ (\Delta \bar{x})^2 = \sum_i \left( \frac{\partial \bar{x}}{\partial x_i} \Delta x_i \right)^2. $$ In the present case, all of the $\Delta x_i$ values are the same, and this works out to be $$ \Delta \bar{x} = \sqrt{ 5 \cdot \left(\frac{1}{5} \Delta x \right)^2} = \frac{\Delta x}{\sqrt{5}} \approx 0.0223 \, \text{mL}. $$ Note that this implies that as as you take more and more repeated measurements, you should get a better and better idea of what the "real" value is, since $\Delta \bar{x}$ is decreasing with time.

That said, there are some caveats to this technique. In particular, it makes a pretty strong (if standard) assumption about your measurement technique: namely, that you're equally likely to get a value that's too high or too low, and that your errors are distributed in some sort of bell curve about some "true" value. The method described here still holds (I think) under certain relaxations of the latter assumption. The former assumption, however, is pretty necessary for this logic to work; and it's easy to think of a situation where it might not hold. (In particular, "last-digit" measurement error might tend to be a bit too high or too low depending on many factors.) Proper error analysis requires a lot of careful thought; there's no one-size-fits-all approach to it.

If we assume that each of the trials had a normally distributed standard error of 0.05 mL, and that all of these errors were independent of each other, then the correct way to combine the errors is via quadrature. The idea here is that the average $\bar{x}$ will depend on each of the measurements $x_1, x_2, \dots$ by $$ \bar{x} = \frac{1}{5} (x_1 + x_2 + x_3 + x_4 + x_5). $$ If each $x_i$ has an uncertainty $\Delta x_i$, then the general formula for the uncertainty in the derived quantity $\bar{x}$ will be $$ (\Delta \bar{x})^2 = \sum_i \left( \frac{\partial \bar{x}}{\partial x_i} \Delta x_i \right)^2. $$ In the present case, all of the $\Delta x_i$ values are the same, and this works out to be $$ \Delta \bar{x} = \sqrt{ 5 \cdot \frac{1}{25} (\Delta x)^2} = \frac{\Delta x}{\sqrt{5}} \approx 0.0223 \, \text{mL}. $$ Note that this implies that as as you take more and more repeated measurements, you should get a better and better idea of what the "real" value is, since $\Delta \bar{x}$ is decreasing with time.

That said, there are some caveats to this technique. In particular, it makes a pretty strong (if standard) assumption about your measurement technique: namely, that you're equally likely to get a value that's too high or too low, and that your errors are statistically distributed about some "true" value. The method described here still holds (I think) under certain relaxations of the latter assumption. The former assumption, however, is pretty necessary for this logic to work; and it's easy to think of a situation where it might not hold. (In particular, "last-digit" measurement error might tend to be a bit too high or too low depending on many factors.) Proper error analysis requires a lot of careful thought; there's no one-size-fits-all approach to it.

If we assume that each of the trials had a normally distributed standard error of 0.05 mL, and that all of these errors were independent of each other, then the correct way to combine the errors is via quadrature. The idea here is that the average $\bar{x}$ will depend on each of the measurements $x_1, x_2, \dots$ by $$ \bar{x} = \frac{1}{5} (x_1 + x_2 + x_3 + x_4 + x_5). $$ If each $x_i$ has an uncertainty $\Delta x_i$, then the general formula for the uncertainty in the derived quantity $\bar{x}$ will be $$ (\Delta \bar{x})^2 = \sum_i \left( \frac{\partial \bar{x}}{\partial x_i} \Delta x_i \right)^2. $$ In the present case, all of the $\Delta x_i$ values are the same, and this works out to be $$ \Delta \bar{x} = \sqrt{ 5 \cdot \frac{1}{25} (\Delta x)^2} = \frac{\Delta x}{\sqrt{5}} \approx 0.0223 \, \text{mL}. $$ Note that this implies that as as you take more and more repeated measurements, you should get a better and better idea of what the "real" value is, since $\Delta \bar{x}$ is decreasing with time.

That said, there are some caveats to this technique. In particular, it makes a pretty strong (if standard) assumption about your measurement technique: namely, that you're equally likely to get a value that's too high or too low, and that your errors are distributed in some sort of bell curve about some "true" value. The method described here still holds (I think) under certain relaxations of the latter assumption. The former assumption, however, is pretty necessary for this logic to work; and it's easy to think of a situation where it might not hold. (In particular, "last-digit" measurement error might tend to be a bit too high or too low depending on many factors.) Proper error analysis requires a lot of careful thought; there's no one-size-fits-all approach to it.