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.