Final answer:
The probability of selecting a person with a temperature greater than 98.6 degrees Fahrenheit from a population with normally distributed body temperatures is 0.5000.
Step-by-step explanation:
The question asks for the probability of selecting a person at random from a population with body temperatures that are normally distributed with a mean of 98.2 and a standard deviation of 0.62. To find this probability, we need to calculate the z-score and look it up in the standard normal distribution table.
First, we calculate the z-score: z = (x - mu) / sigma, where x is the body temperature, mu is the mean, and sigma is the standard deviation. In this case, x is 98.6, mu is 98.2, and sigma is 0.62. Plugging in the values, we get z = (98.6 - 98.2) / 0.62 = 0.64.
Next, we look up the z-score in the standard normal distribution table. The closest value to 0.64 is 0.7389. Since we're looking for the probability of selecting a person with a temperature greater than 98.6, we subtract the probability of selecting a person with a temperature less than or equal to 98.6 from 0.5 (since the distribution is symmetric).
Therefore, the answer is (c) 0.5000.