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?

This is a wonderful page for your type of question

http://davidmlane.com/hyperstat/z_table.html

You can use the data directly in the first applet, and in the second use z-scores.

b) z-score for 99
= (99-98.2)/.62 = 1.29

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 can use the z-score formula:

Z = (X - μ) / σ

Where X is the value, μ is the mean, and σ is the standard deviation.

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

To find the percentile score, we can refer to the z-table or use a statistical calculator. For a z-score of 1.29, the percentile score is approximately 90.02%. Therefore, if your body temperature is 99.00 °F, your percentile score is approximately 90.02%.

b. To convert 99.00 °F to a standard score (z-score), we can use the same formula:

Z = (X - μ) / σ

Where X is the value, μ is the mean, and σ is the standard deviation.

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

Therefore, the standard score (z-score) for a body temperature of 99.00 °F is 1.29.

c. A body temperature of 99.00 °F is not considered unusual. This is because it falls within the normal range of body temperatures for healthy adults. The data provided states that the mean body temperature is 98.20 °F, so a temperature of 99.00 °F is slightly higher but still within a reasonable range.

d. To find the likelihood that the mean of fifty adults' body temperatures is 97.98 °F or lower, we need to find the z-score for 97.98 °F and calculate the area under the normal distribution curve.

Z = (X - μ) / (σ / sqrt(n))

Where X is the value, μ is the mean, σ is the standard deviation, and n is the sample size.

Z = (97.98 - 98.20) / (0.62 / sqrt(50))
Z = -2.08

We can refer to the z-table or use a statistical calculator to find the probability associated with a z-score of -2.08. The probability is approximately 0.0188, or 1.88%.

Therefore, the likelihood that the mean of fifty adults' body temperatures is 97.98 °F or lower is approximately 1.88%.

e. A person's body temperature of 101.00 °F may be considered unusual. This is because it is above the normal range of body temperatures for healthy adults. The data provided states that the mean body temperature is 98.20 °F, and a temperature of 101.00 °F is significantly higher than that. It could potentially indicate a fever or an underlying medical condition. Further assessment and medical attention may be necessary.

f. To find the body temperature corresponding to the 95th percentile, we can use the z-score formula:

Z = (X - μ) / σ

Where Z is the z-score corresponding to the desired percentile, X is the value, μ is the mean, and σ is the standard deviation.

For the 95th percentile, the corresponding z-score is approximately 1.645. We can rearrange the formula to solve for X:

X = Z * σ + μ

X = 1.645 * 0.62 + 98.20
X = 1.0209 + 98.20
X ≈ 99.22 °F

Therefore, the body temperature at the 95th percentile is approximately 99.22 °F.

g. To find the body temperature corresponding to the 5th percentile, we can use the same approach as in the previous question.

For the 5th percentile, the corresponding z-score is approximately -1.645. Using the formula:

X = Z * σ + μ

X = -1.645 * 0.62 + 98.20
X = -1.0219 + 98.20
X ≈ 97.18 °F

Therefore, the body temperature at the 5th percentile is approximately 97.18 °F.

h. To find the percentage of normal and healthy adults with a fever according to the cutoff temperature of 100.6 °F, we can calculate the area under the normal distribution curve above this temperature.

Z = (X - μ) / σ

Where X is the value (100.6 °F), μ is the mean (98.20 °F), and σ is the standard deviation (0.62 °F).

Z = (100.6 - 98.20) / 0.62
Z = 3.87

The z-table or a statistical calculator can be used to find the probability associated with a z-score of 3.87. The probability is approximately 0.9999, or 99.99%.

Therefore, according to the cutoff of 100.6 °F, approximately 99.99% of normal and healthy adults would be considered to have a fever. This percentage suggests that the cutoff of 100.6 °F may not be appropriate for distinguishing between febrile and non-febrile conditions in normal and healthy adults. Further evaluation by medical professionals is necessary to determine the appropriate temperature cutoff for fever.

a. To find the percentile score for a body temperature of 99.00 °F, we first need to convert it into a z-score. The formula to calculate the z-score is:

z = (x - μ) / σ

where x is the value (99.00 °F), μ is the mean (98.20 °F), and σ is the standard deviation (0.62 °F).

Plugging in the values, we get:

z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z ≈ 1.29

Once we have the z-score, we can use a standard normal distribution table or a statistical software to find the corresponding percentile. From the table, a z-score of 1.29 corresponds to a percentile score of approximately 90.92%.

Therefore, if your body temperature is 99.00 °F, your percentile score is approximately 90.92%.

b. The standard score, also known as the z-score, measures how many standard deviations a data point is from the mean. We already calculated the z-score in part a, which is approximately 1.29.

Therefore, if your body temperature is 99.00 °F, your z-score is approximately 1.29.

c. To determine if a body temperature of 99.00 °F is unusual, we need to consider the standard score (z-score). In a normal distribution, a z-score between -2 and +2 is generally considered within the normal range.

In this case, the z-score of 1.29 falls within this range, indicating that a body temperature of 99.00 °F is not unusual.

d. To find the likelihood that the mean of the body temperatures of fifty randomly selected adults is 97.98 °F or lower, we need to use the sampling distribution of means.

The mean of the sampling distribution of means will be equal to the mean of the original population, which is 98.20 °F.

The standard deviation of the sampling distribution of means, also known as the standard error of the mean, is equal to the standard deviation of the original population divided by the square root of the sample size. In this case, the sample size is fifty.

σ (sampling distribution of means) = σ (original population) / √n

σ (sampling distribution of means) = 0.62 / √50

Once we have the standard deviation of the sampling distribution of means, we can find the z-score for 97.98 °F using the formula from part a and then use a standard normal distribution table or a statistical software to find the corresponding probability.

e. To determine if a body temperature of 101.00 °F is unusual, we can calculate its z-score using the formula from part b. Let's do that:

z = (101.00 - 98.20) / 0.62
z ≈ 4.52

The z-score of 4.52 indicates that a body temperature of 101.00 °F is very unlikely to occur in a normal distribution. Therefore, it is considered unusual.

Based on this information, one could conclude that a body temperature of 101.00 °F may indicate an abnormality or illness and further medical attention may be required.

f. To find the body temperature corresponding to the 95th percentile, we need to find the z-score that corresponds to a cumulative area of 0.95 in a standard normal distribution table.

Looking up the value in the table, we find that a z-score of 1.645 corresponds to a cumulative area of 0.95.

Using the formula from part a, we can solve for x (body temperature):

1.645 = (x - 98.20) / 0.62

Solving for x, we get:

x - 98.20 = 1.645 * 0.62
x - 98.20 = 1.0199
x ≈ 99.22

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

g. To find the body temperature corresponding to the 5th percentile, we need to find the z-score that corresponds to a cumulative area of 0.05 in a standard normal distribution table.

Looking up the value in the table, we find that a z-score of -1.645 corresponds to a cumulative area of 0.05.

Using the formula from part a, we can solve for x (body temperature):

-1.645 = (x - 98.20) / 0.62

Solving for x, we get:

x - 98.20 = -1.0199
x ≈ 97.18

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

h. To determine what percentage of normal and healthy adults would be considered to have a fever according to Bellevue Hospital's cutoff of 100.6 °F, we can calculate the proportion of individuals with body temperatures above that cutoff using the normal distribution.

We need to find the cumulative area to the right of the cutoff temperature using the z-score formula:

z = (x - μ) / σ

Plugging in the values:

z = (100.6 - 98.20) / 0.62
z ≈ 3.87

Looking up the cumulative area for a z-score of 3.87 in the standard normal distribution table, we find that it is very close to 1 (almost the entire area under the curve).

Therefore, a very small percentage of normal and healthy adults would be considered to have a fever using Bellevue Hospital's cutoff of 100.6 °F.

This suggests that the cutoff may not be appropriate for determining fever in normal and healthy adults, as it would classify almost all of them as having a fever.