The daily leisure activity time of residents in a certain region follows a normal distribution with mean $m$ minutes and standard deviation $\sigma$ minutes. When 16 residents are randomly sampled and the sample mean of daily leisure activity time is 75 minutes, the 95\% confidence interval for the population mean $m$ is $a \leq m \leq b$. When 16 residents are randomly sampled again and the sample mean of daily leisure activity time is 77 minutes, the 99\% confidence interval for the population mean $m$ is $c \leq m \leq d$. Find the value of $\sigma$ that satisfies $d - b = 3.86$. (Here, when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95 , \mathrm { P } ( | Z | \leq 2.58 ) = 0.99$.) [4 points]
The daily leisure activity time of residents in a certain region follows a normal distribution with mean $m$ minutes and standard deviation $\sigma$ minutes. When 16 residents are randomly sampled and the sample mean of daily leisure activity time is 75 minutes, the 95\% confidence interval for the population mean $m$ is $a \leq m \leq b$. When 16 residents are randomly sampled again and the sample mean of daily leisure activity time is 77 minutes, the 99\% confidence interval for the population mean $m$ is $c \leq m \leq d$.\\
Find the value of $\sigma$ that satisfies $d - b = 3.86$.\\
(Here, when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95 , \mathrm { P } ( | Z | \leq 2.58 ) = 0.99$.) [4 points]