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 $S_{best} = ({\sum_{i = 1}^{k} {1 \over S_i^2}})^{-1/2}$.

Note: Sample values (e.g., mean and standard deviation) are best-estimates of the unknown population values.

(For example, see the text Data Analysis for Scientists and Engineers by Meyer for details.)

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.)