Assume the body temperture of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F...

a. If you have a body temperture of 99.00 degrees F, what is your percentile score?

b. Convert 99.00 degrees F to a standard score (or a z-score).

c. Is a body temperature of 99.00 degrees F unusual? Why or why not?

d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperature is 97.98 degrees F or lower?

e. A person's body temperature is found to 101.00 degrees F. Is the result unusual? Why or why not?

f. What body temperature is the 95th percentile?

g. 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 proportions corresponding to your Z scores.

d. Z = (mean1 - mean2)/Standard Error (SE) of the mean

SE = SD/√n

This information should lead you to find your own answers. We do not do your work for you.

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?

a 101 body temperature is unusual

a. To find the percentile score, we need to determine the area under the normal distribution curve to the left of the given value. We can use the z-score formula to calculate this area.

First, we find the z-score:
z = (x - μ) / σ
where x is the given value (99.00 degrees F), μ is the mean (98.20 degrees F), and σ is the standard deviation (0.62 degrees F).

Plugging in the values, we get:
z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z ≈ 1.29

Next, we look up the z-score on a standard normal distribution table or use a calculator to find the area to the left of z = 1.29. This gives us the percentile score.

Using a standard normal distribution table, we find that the area to the left of z = 1.29 is approximately 0.9015. To convert this to a percentile score, we multiply by 100:
Percentile score = 0.9015 * 100 ≈ 90.15%

Therefore, if you have a body temperature of 99.00 degrees F, your percentile score is approximately 90.15%.

b. To convert 99.00 degrees F to a standard score (or a z-score), we use the formula:
z = (x - μ) / σ
where x is the given value (99.00 degrees F), μ is the mean (98.20 degrees F), and σ is the standard deviation (0.62 degrees F).

Plugging in the values, we get:
z = (99.00 - 98.20) / 0.62
z = 0.80 / 0.62
z ≈ 1.29

Therefore, 99.00 degrees F corresponds to a z-score of approximately 1.29.

c. To determine if a body temperature of 99.00 degrees F is unusual, we can compare it to the normal distribution. In general, values that fall within 2 standard deviations of the mean (i.e., between μ - 2σ and μ + 2σ) are considered typical or common. Anything outside this range can be considered unusual.

Since the given body temperature of 99.00 degrees F falls within 2 standard deviations of the mean, it is not considered unusual.

d. To find the likelihood that the mean of the body temperature of fifty randomly selected adults is 97.98 degrees F or lower, we use the Central Limit Theorem.

The Central Limit Theorem states that for a large enough sample size, the distribution of sample means will approach a normal distribution regardless of the shape of the population distribution.

In this case, we need to find the z-score for 97.98 degrees F using the formula:
z = (x - μ) / (σ / sqrt(n))
where x is the given value (97.98 degrees F), μ is the population mean (98.20 degrees F), σ is the population standard deviation (0.62 degrees F), and n is the sample size (50 adults).

Plugging in the values, we get:
z = (97.98 - 98.20) / (0.62 / sqrt(50))
z ≈ -0.22 / 0.08795
z ≈ -2.5

Next, we look up the z-score on a standard normal distribution table or use a calculator to find the area to the left of z = -2.5. This gives us the likelihood.

Using a standard normal distribution table, we find that the area to the left of z = -2.5 is approximately 0.00621.

Therefore, the likelihood that the mean of the body temperature of fifty randomly selected adults is 97.98 degrees F or lower is approximately 0.00621 (or 0.621%).

e. To determine if a body temperature of 101.00 degrees F is unusual, we can compare it to the normal distribution. In general, values that fall within 2 standard deviations of the mean (i.e., between μ - 2σ and μ + 2σ) are considered typical or common. Anything outside this range can be considered unusual.

Since the given body temperature of 101.00 degrees F falls outside 2 standard deviations of the mean, it is considered unusual.

f. To find the body temperature at the 95th percentile, we need to find the z-score that corresponds to the desired percentile. We can then use the z-score formula to calculate the corresponding body temperature.

First, we need to find the z-score that corresponds to the 95th percentile. Using a standard normal distribution table or a calculator, we find that the z-score for the 95th percentile is approximately 1.645.

Next, we rearrange the z-score formula:
z = (x - μ) / σ
to solve for x:
x = μ + (z * σ)

Plugging in the values, we get:
x = 98.20 + (1.645 * 0.62)
x = 98.20 + 1.0189
x ≈ 99.22 degrees F

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

g. To find the body temperature at the 5th percentile, we follow a similar process as in the previous question.

First, we need to find the z-score that corresponds to the 5th percentile. Using a standard normal distribution table or a calculator, we find that the z-score for the 5th percentile is approximately -1.645.

Using the z-score formula, we can calculate the corresponding body temperature:
x = μ + (z * σ)
x = 98.20 + (-1.645 * 0.62)
x = 98.20 - 1.0189
x ≈ 97.18 degrees F

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