Suppose that the TSH (Thyroid Stimulating Hormone) levels among healthy individuals are normally distributed with a mean of $3.2 \frac{\text { units }}{\mathrm{mL}}$. Suppose also that exactly $95 \%$ of healthy individuals have TSH levels below $5.8 \frac{\text { units }}{\mathrm{mL}}$. Find the standard deviation of the distribution of TSH levels of healthy individuals. Carry your intermedlate computations to at least four decimal places. Round your answer to at least two decimal places.

Step 1 ：We are given that the TSH levels are normally distributed with a mean of 3.2 units/mL.

Step 2 ：We are also given that 95% of healthy individuals have TSH levels below 5.8 units/mL. This means that 5.8 units/mL is the 95th percentile of the distribution.

Step 3 ：In a standard normal distribution, the 95th percentile corresponds to a z-score of approximately 1.645. The z-score is calculated as \((X - μ) / σ\), where X is the value from the dataset, μ is the mean, and σ is the standard deviation.

Step 4 ：We can set up the equation \(1.645 = (5.8 - 3.2) / σ\) and solve for σ to find the standard deviation of the distribution.

Step 5 ：By solving the equation, we find that the standard deviation of the distribution of TSH levels of healthy individuals is approximately \(\boxed{1.58}\) units/mL.