2. Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Maryland researchers).

a. If you have a body temperature of 99.00 °F, what is your percentile score?
b. Convert 99.00 °F to a standard score (or a z-score).
c. Is a body temperature of 99.00 °F unusual? Why or why not?
d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?
e. A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not? What should you conclude?
f. What body temperature is the 95th percentile?
g. What body temperature is the 5th percentile?
h. Bellevue Hospital in New York City uses 100.6 °F as the lowest temperature considered to indicate a fever. What percentage of normal and healthy adults would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6 °F is appropriate?

a. To find the percentile score for a body temperature of 99.00 °F, we need to calculate the z-score and then use a standard normal distribution table. The formula for z-score is:

z = (x - μ) / σ

where x is the individual value, μ is the mean, and σ is the standard deviation.

In this case:
x = 99.00 °F
μ = 98.20 °F
σ = 0.62 °F

So, the z-score is:
z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z ≈ 1.29

Using a standard normal distribution table or calculator, we can find the percentile corresponding to a z-score of 1.29. The percentile score is approximately 90.64%. Therefore, a body temperature of 99.00 °F corresponds to a percentile score of about 90.64%.

b. To convert 99.00 °F to a standard score or z-score, we use the same formula as before:

z = (x - μ) / σ

where x is the individual value, μ is the mean, and σ is the standard deviation.

In this case:
x = 99.00 °F
μ = 98.20 °F
σ = 0.62 °F

So, the z-score is:
z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z ≈ 1.29

Therefore, a body temperature of 99.00 °F corresponds to a z-score of approximately 1.29.

c. To determine if a body temperature of 99.00 °F is unusual, we can compare the z-score to a standard normal distribution. Generally, z-scores greater than 2 or less than -2 are considered unusual.

In this case, the z-score is approximately 1.29, which is within the normal range of -2 to 2. Therefore, a body temperature of 99.00 °F is not considered unusual.

d. To find the likelihood that the mean of 50 randomly selected adults' body temperatures is 97.98 °F or lower, we need to use the Central Limit Theorem. The Central Limit Theorem states that the distribution of sample means is approximately normally distributed, regardless of the underlying population distribution, when the sample size is sufficiently large.

Given:
n (sample size) = 50
μ (population mean) = 98.20 °F
σ (population standard deviation) = 0.62 °F
x (sample mean) = 97.98 °F

We can calculate the standard error (SE) using the formula:

SE = σ / √n

In this case:
SE = 0.62 / √50
SE ≈ 0.088

Now, we can calculate the z-score for the sample mean using the formula:

z = (x - μ) / SE

z = (97.98 - 98.20) / 0.088
z ≈ -0.22 / 0.088
z ≈ -2.50

Using a standard normal distribution table or calculator, we find that the likelihood of obtaining a z-score of -2.50 or lower is approximately 0.0062, or 0.62%. Therefore, the likelihood that the mean of 50 adults' body temperatures is 97.98 °F or lower is about 0.62%.

e. To determine if a body temperature of 101.00 °F is unusual, we can compare the z-score to a standard normal distribution. Generally, z-scores greater than 2 or less than -2 are considered unusual.

To calculate the z-score:
x = 101.00 °F
μ = 98.20 °F
σ = 0.62 °F

z = (101.00 - 98.20) / 0.62
z = 2.8 / 0.62
z ≈ 4.516

The z-score of approximately 4.516 is greater than 2, indicating that a body temperature of 101.00 °F is considered unusual.

We can conclude that a body temperature of 101.00 °F is unusual, as it falls outside the normal range.

f. To find the body temperature that corresponds to the 95th percentile, we need to use a standard normal distribution table or calculator. The 95th percentile corresponds to a z-score of approximately 1.645.

Using the formula:

z = (x - μ) / σ

where x is the body temperature, μ is the mean, σ is the standard deviation, and z is the z-score.

In this case, we have:
μ = 98.20 °F
σ = 0.62 °F
z = 1.645

To solve for x (the body temperature), we rearrange the formula:
x = z * σ + μ

x = 1.645 * 0.62 + 98.20
x ≈ 1.0199 + 98.20
x ≈ 99.2199

Therefore, the body temperature that corresponds to the 95th percentile is approximately 99.22 °F.

g. To find the body temperature that corresponds to the 5th percentile, we use the same process as in the previous step. The 5th percentile corresponds to a z-score of approximately -1.645.

Using the formula:
z = (x - μ) / σ

where x is the body temperature, μ is the mean, σ is the standard deviation, and z is the z-score.
μ = 98.20 °F
σ = 0.62 °F
z = -1.645

To solve for x:
x = z * σ + μ

x = -1.645 * 0.62 + 98.20
x ≈ -1.0199 + 98.20
x ≈ 97.1801

Therefore, the body temperature that corresponds to the 5th percentile is approximately 97.18 °F.

h. To determine the percentage of normal and healthy adults who would be considered to have a fever according to Bellevue Hospital's cutoff of 100.6 °F, we need to calculate the z-score based on this cutoff.

Using the formula:
z = (x - μ) / σ

where x is the cutoff temperature, μ is the mean, σ is the standard deviation, and z is the z-score.
x = 100.6 °F
μ = 98.20 °F
σ = 0.62 °F

z = (100.6 - 98.20) / 0.62
z ≈ 2.459

To find the percentage of normal and healthy adults with a body temperature greater than 100.6 °F, we can use the standard normal distribution table or calculator to find the area to the right of the z-score of 2.459.

The area to the right of 2.459 is approximately 0.0067, or 0.67%.

Therefore, approximately 0.67% of normal and healthy adults would be considered to have a fever based on Bellevue Hospital's cutoff of 100.6 °F.

This low percentage indicates that the cutoff of 100.6 °F may not be appropriate for determining if someone has a fever in the general population of normal and healthy adults.