Healthy people have body temperatures that are normally distributed with a mean of 98.20∘F and a standard deviation of 0.62∘F . (a) If a healthy person is randomly selected, what is the probability that he or she has a temperature above 98.9∘F?

This is a question of finding the probability of a normally distributed variable, and for this, we have to know that the normal distribution is determined by two parameters: the population mean and the population standard deviation. In this case, they are, respectively, [tex] \\ \mu = 98.20[/tex] Fahrenheit degrees and [tex] \\ \sigma = 0.62[/tex] Fahrenheit degrees.

To find probabilities, we can "transform" these "raw" scores into z-scores, or standardized values, using the z-score formula. After this, we can consult the cumulative standard normal table (available in Statistics books or on the Internet) using this z-score (e.g., z = a), for which we have the corresponding cumulative probability, that is, P(z<a). Of course, we can also use statistical software, or even a spreadsheet, to find such probabilities.

A z-score tells us the distance from the mean in standard deviations units for the standardized value of the raw score. A positive value indicates that the value is above the mean. Conversely, a negative value tells us that the value is below the mean.

The formula for this z-score is as follows

z = [tex] \\ \frac{x - \mu}{\sigma}[/tex] [1]

Where x is the raw score (x = 98.9 Fahrenheit degrees, in this case).

With this information at hand, we can solve the question.

The probability that a healthy person has a temperature above 98.9 Fahrenheit degrees

We need to find P(x>98.9).

Using formula [1], the z-score for x = 98.9 is

z = [tex] \\ \frac{x - \mu}{\sigma}[/tex]

z = [tex] \\ \frac{98.9 - 98.20}{0.62}[/tex]

z = [tex] \\ \frac{0.70}{0.62}[/tex]

z = [tex] \\ 1.12903 \approx 1.13[/tex]

With this value for z (z = 1.13), we can consult the cumulative standard normal table to find the probability for z = 1.13 or P(z<1.13), for which we have a value of P(z<1.13) = 0.8708.

However, we are asked for P(x>98.9) = P(z>1.13), which is the complement probability for P(z<1.13) or

[tex] \\ P(x>98.9) = P(z>1.13) = 1 - P(z<1.13)[/tex]

[tex] \\ P(x>98.9) = P(z>1.13) = 1 - 0.8708[/tex]

[tex] \\ P(x>98.9) = P(z>1.13) = 0.1292[/tex]

Remember that z is the standardized value for x, so P(x>98.9) = P(z>1.13).

Therefore, the probability that a healthy person has a temperature above 98.9 Fahrenheit degrees is about [tex] \\ P(x>98.9) = P(z>1.13) = 0.1292[/tex] or 12.92%.

We can see this area in the graph below.