1. The problem is to find the standard deviation $\sigma$ from given confidence intervals.\n2. Confidence intervals are typically given as $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level, $n$ is the sample size, and $\sigma$ is the population standard deviation.\n3. From the confidence interval $[L,U]$, the margin of error (ME) is $ME = \frac{U-L}{2}$.\n4. The margin of error relates to the standard deviation by $ME = z \frac{\sigma}{\sqrt{n}}$.\n5. Rearranging for $\sigma$ gives $$\sigma = \frac{ME \times \sqrt{n}}{z} = \frac{(U-L)/2 \times \sqrt{n}}{z}.$$\n6. So, given the confidence interval bounds $L$ and $U$, the confidence level (to find $z$), and the sample size $n$, you can compute the standard deviation $\sigma$.\n7. Note: This method assumes the confidence interval is based on a normal distribution and known $z$. For small samples or unknown population variance, other methods apply.\n\nFinal formula: $$\sigma = \frac{(U-L)}{2} \times \frac{\sqrt{n}}{z}.$$