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?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.

d. For distribution of mean rather than scores, use SEm instead of SD. SEm = SD/√(n-1) Then use same table.

a. To find the percentile score for a body temperature of 99.00 °F, we need to calculate the area under the normal distribution curve to the left of 99.00 °F.

First, we need to standardize the value by subtracting the mean and dividing by the standard deviation.

Z-score = (99.00 - 98.20) / 0.62 = 1.29

Next, we can use a z-score table or a statistical software to find the area under the curve to the left of a z-score of 1.29. The area represents the percentile score.

The percentile score for a body temperature of 99.00 °F is approximately 90.46%.

b. To convert 99.00 °F to a standard score or z-score, we use the formula:
Z-score = (x - mean) / standard deviation

Z-score = (99.00 - 98.20) / 0.62 = 1.29

The standard score or z-score for a body temperature of 99.00 °F is 1.29.

c. To determine if a body temperature of 99.00 °F is unusual, we need to consider the range of values that fall within a certain number of standard deviations from the mean.

Based on the given mean and standard deviation, we can calculate the range within 2 standard deviations of the mean as follows:

Lower Bound: mean - (2 * standard deviation) = 98.20 - (2 * 0.62) = 97.58 °F
Upper Bound: mean + (2 * standard deviation) = 98.20 + (2 * 0.62) = 98.82 °F

Since 99.00 °F falls within this range, it is not considered unusual based on this definition. However, this is a subjective determination, and the definition of "unusual" may vary depending on the context.

d. To find the likelihood that the mean of the body temperatures of 50 randomly selected adults is 97.98 °F or lower, we need to use the Central Limit Theorem.

With a sample size of 50, the sample mean follows a normal distribution with the same mean as the population (98.20 °F) and a standard deviation of the population divided by the square root of the sample size (0.62 / √50).

We can then find the z-score for 97.98 °F:

Z-score = (97.98 - 98.20) / (0.62 / √50)

Using a z-score table or statistical software, we can find the corresponding probability, which represents the likelihood that the mean of the body temperatures of 50 randomly selected adults is 97.98 °F or lower.

e. To determine if a body temperature of 101.00 °F is unusual, we follow the same process as in step c.

Calculate the range within 2 standard deviations of the mean:
Lower Bound: mean - (2 * standard deviation) = 98.20 - (2 * 0.62) = 97.58 °F
Upper Bound: mean + (2 * standard deviation) = 98.20 + (2 * 0.62) = 98.82 °F

Since 101.00 °F exceeds the upper bound of this range, it can be considered unusual.

To conclude, a body temperature of 101.00 °F may indicate a deviation from the normal range and could potentially suggest a fever or health issue.

f. To find the body temperature corresponding to the 95th percentile, we need to calculate the z-score that corresponds to a cumulative probability of 0.95.

Using a z-score table or statistical software, we can find the z-score that corresponds to a cumulative probability of 0.95.

Z-score = 1.645 (approx.)

Next, we can convert the z-score back to the original measurement scale using the formula:

Body Temperature = (Z-score * standard deviation) + mean

Body Temperature = (1.645 * 0.62) + 98.20

The body temperature at the 95th percentile is approximately 99.59 °F.

g. To find the body temperature corresponding to the 5th percentile, we follow the same process as in step f.

Using a z-score table or statistical software, we can find the z-score that corresponds to a cumulative probability of 0.05.

Z-score = -1.645 (approx.)

Converting the z-score back to the original measurement scale:

Body Temperature = (Z-score * standard deviation) + mean

Body Temperature = (-1.645 * 0.62) + 98.20

The body temperature at the 5th percentile is approximately 96.76 °F.

h. To determine the percentage of normal and healthy adults that would be considered to have a fever according to Bellevue Hospital's cutoff of 100.6 °F, we need to find the area under the curve to the right of 100.6 °F.

First, we standardize the value by subtracting the mean and dividing by the standard deviation:

Z-score = (100.6 - 98.20) / 0.62

Using a z-score table or statistical software, we can find the area to the right of the z-score.

The percentage of normal and healthy adults with a body temperature of 100.6 °F or higher can then be calculated by subtracting the area under the curve to the left of the z-score from 1 (or multiplying the area by 100%).

Whether the percentage suggests that a cutoff of 100.6 °F is appropriate would depend on additional factors such as the specific population being considered and any relevant medical guidelines or research on fever thresholds.