We need to be careful here. Suppose the data you report in your question are means and standard deviations of the means for a series of random samples. That is, $1.510 \pm 0.085$ is the mean $\pm$ the standard deviation of the mean for one sample consisting of a number of individual measurements, 1.608±0.089 is the mean ± the standard deviation of the mean for another sample consisting of a number of individual measurements, and so on.
Consider a series of $i = 1, 2, ...,k$ random samples, the $i^{th}$ sample consisting of $n_i$ specific values for the random variable of concern. Each sample can have a different number of specific values. For each sample you evaluate and report the mean and the standard deviation of the mean. The mean of the $i^{th}$ sample is $m_i = {1 \over n_{i}} \sum_{j}^{n_i} y_{ji}$ where $y_{ji}$ is the $j^{th}$ value in the $i^{th}$ sample. The standard deviation of the mean for the $i^{th}$ sample is $S_i = \sqrt{s_i^2 \over n_i}$ where $s_i = \sqrt{{\sum_{j}^{n_i} (y_{ji} - m_i)^2} \over {n_i - 1} }$ is the standard deviation for the $i^{th}$ sample.
The best estimate for the mean is $ m_{best} = {\sum_{i = 1}^{k} m_i/S_i^2 \over \sum_{i = 1}^{k} {1 \over S_i^2}}$ and the best estimate for the standard deviation of the mean is $S_{best} = ({\sum_{i = 1}^{k} {1 \over S_i^2}})^{-1/2}$. You report $m_{best} \pm S_{best}$ for your final result.
Note: The sample values mean $m_{best}$ and standard deviation $S_{best}$ are best-estimates for the unknown population values mean $\mu$ and standard deviation of the mean $\sigma_{\mu}$.
(For example, see the text Data Analysis for Scientists and Engineers by Meyer for details.)