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. (0.1 point) If you have a body temperature of 99.00 °F, what is your percentile score?

b. (0.1 point) Convert 99.00 °F to a standard score (or a z-score).



c. 1. (0.1 point) Is a body temperature of 99.00 °F unusual?



2. (0.1 point) Why or why not?



d. (0.1 point) Fifty adults are randomly selected. What is the likelihood that the mean

of their body temperatures is 97.98 °F or lower?



e. (0.1 point) A person’s body temperature is found to be 101.00 °F. Is the result unusual?



(0.1 point) Why or Why Not and What should you conclude?









f. (0.1 point) What body temperature is the 95th percentile? :



g. (0.1 point) What body temperature is the 5th percentile?

Z = (x - mean)/SD

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

You can also reverse the process to get the Z score for a percentile and plug that in the above formula to get the temperature.

Make your own conclusions.

a. To find the percentile score for a body temperature of 99.00 °F, we need to calculate the z-score first and then use a z-table.

The z-score is calculated using the formula:
z = (x - mean) / standard deviation

In this case, the mean is 98.20 °F and the standard deviation is 0.62 °F. Substituting these values with the given body temperature of 99.00 °F into the formula:
z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z = 1.29

Now we can use the z-score to find the percentile score from the z-table. The z-score of 1.29 corresponds to approximately 90.85% on the z-table. Therefore, a body temperature of 99.00 °F has a percentile score of approximately 90.85%.

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

Substituting the values:
z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z = 1.29

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

c. 1. To determine if a body temperature of 99.00 °F is unusual, we need to compare it to the normal distribution of body temperatures.

2. Usually, we consider a body temperature outside the range of two standard deviations from the mean as unusual or abnormal. In our case, the mean is 98.20 °F and the standard deviation is 0.62 °F.

Two standard deviations from the mean would be:
Lower limit = mean - (2 * standard deviation) = 98.20 - (2 * 0.62) = 96.96 °F
Upper limit = mean + (2 * standard deviation) = 98.20 + (2 * 0.62) = 99.44 °F

Since a body temperature of 99.00 °F falls within the range of two standard deviations from the mean, it is not considered unusual.

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

The mean of the sampling distribution is equal to the population mean (98.20 °F), and the standard deviation of the sampling distribution (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size (0.62 °F / √50).

Next, we calculate the z-score using the formula:
z = (sample mean - population mean) / standard error

Substituting the values:
z = (97.98 - 98.20) / (0.62 / √50) = -0.22 / (0.0878) = -2.51

Using the z-table, a z-score of -2.51 corresponds to a percentile of approximately 0.0063 or 0.63%.

Therefore, the likelihood that the mean of fifty adults' body temperatures is 97.98 °F or lower is approximately 0.63%.

e. To determine if a body temperature of 101.00 °F is unusual, we need to compare it to the normal distribution of body temperatures.

Usually, we consider a body temperature outside the range of two standard deviations from the mean as unusual or abnormal. In this case, a body temperature of 101.00 °F falls outside the upper limit of two standard deviations from the mean.

The upper limit is mean + (2 * standard deviation) = 98.20 + (2 * 0.62) = 99.44 °F.

Since 101.00 °F is outside this range, it is considered unusual.

We should conclude that a body temperature of 101.00 °F is higher than normal and may indicate a fever or some other health condition.

f. To find the body temperature at the 95th percentile, we need to use the z-score and the z-table.

In the z-table, the 95th percentile corresponds to a z-score of approximately 1.645.

Now we can use the formula to calculate the body temperature:
z = (x - mean) / standard deviation

Rearranging the formula and substituting the values:
x = (z * standard deviation) + mean
x = (1.645 * 0.62) + 98.20
x = 1.0219 + 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 at the 5th percentile, we follow a similar process as in part f.

In the z-table, the 5th percentile corresponds to a z-score of approximately -1.645.

Using the formula:
x = (z * standard deviation) + mean
x = (-1.645 * 0.62) + 98.20
x = -1.0209 + 98.20
x ≈ 97.18 °F

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