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?
Use the processes from your previous post.
However for d, you are using a distribution of means rather than Individual scores.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
That table can be used for your other problems too, finding Z scores to convert from percentile to score.
Am I close? a- 99%, b- 1.29, c- yes, unusual because only 1% of the healthy adult population would be higher,d- 24.5 of the 50 are to be at or below 98.98, e- 4.52, an unusual result and the adult is probably not healthy because the 101 is outside the niormal range by a large percentage. f- 98.97
a. To find the percentile score of a body temperature of 99.00 °F, we need to use the standard normal distribution. First, we calculate the z-score, which represents the number of standard deviations a particular value is from the mean. The formula for the z-score is:
z = (x - μ) / σ
where:
x = body temperature (99.00 °F)
μ = mean (98.20 °F)
σ = standard deviation (0.62 °F)
Plugging in the values, we get:
z = (99.00 - 98.20) / 0.62 = 1.29
To find the percentile score associated with a z-score of 1.29, we can use a standard normal distribution table or a calculator. In this case, the percentile score is approximately 90.65%. This means that a body temperature of 99.00 °F is at the 90.65th percentile.
b. To convert 99.00 °F to a standard score (z-score), we can use the same formula as in part a:
z = (x - μ) / σ
Plugging in the values:
z = (99.00 - 98.20) / 0.62 = 1.29
So, a body temperature of 99.00 °F has a z-score of 1.29.
c. To determine if a body temperature of 99.00 °F is unusual, we can compare its z-score to the standard normal distribution. Typically, z-scores greater than 2 or less than -2 are considered unusual. In this case, the z-score of 1.29 is less than 2, so a body temperature of 99.00 °F would not be considered unusually high.
d. To find the likelihood that the mean of the body temperatures of fifty adults is 97.98 °F or lower, we need to calculate the z-score for this value. The formula remains the same:
z = (x - μ) / (σ / sqrt(n))
where:
x = 97.98 °F
μ = mean (98.20 °F)
σ = standard deviation (0.62 °F)
n = sample size (50)
Plugging in the values:
z = (97.98 - 98.20) / (0.62 / sqrt(50)) ≈ -1.16
Using a standard normal distribution table or a calculator, we can find the probability associated with a z-score of -1.16. The probability is approximately 12.49%. Therefore, the likelihood that the mean of their body temperatures is 97.98 °F or lower is approximately 12.49%.
e. To determine if a body temperature of 101.00 °F is unusual, we can calculate the z-score using the formula:
z = (x - μ) / σ
Plugging in the values:
z = (101.00 - 98.20) / 0.62 ≈ 4.52
A z-score of 4.52 is significantly higher than 2, indicating that a body temperature of 101.00 °F is considered unusual.
We should conclude that the person with a body temperature of 101.00 °F has a significantly high temperature compared to the normal range of healthy adults.
f. To find the body temperature corresponding to the 95th percentile, we need to find the z-score associated with that percentile. Using a standard normal distribution table or calculator, the z-score corresponding to the 95th percentile is approximately 1.645.
We can solve for the temperature using the formula for z-score:
z = (x - μ) / σ
Plugging in the values:
1.645 = (x - 98.20) / 0.62
Solving for x, we get:
x ≈ (1.645 * 0.62) + 98.20 ≈ 99.36 °F
Therefore, the body temperature at the 95th percentile is approximately 99.36 °F.
g. To find the body temperature corresponding to the 5th percentile, we can follow the same steps as in part f, but using the z-score associated with the 5th percentile, which is approximately -1.645.
Using the formula for z-score:
-1.645 = (x - 98.20) / 0.62
Solving for x, we get:
x ≈ (-1.645 * 0.62) + 98.20 ≈ 97.95 °F
Therefore, the body temperature at the 5th percentile is approximately 97.95 °F.
h. To determine the percentage of normal and healthy adults that would be considered to have a fever, we need to find the proportion of individuals with body temperatures higher than 100.6 °F.
We can calculate the z-score for 100.6 °F using the formula:
z = (x - μ) / σ
Plugging in the values:
z = (100.6 - 98.20) / 0.62 ≈ 3.87
A z-score of 3.87 is significantly higher than 2, indicating that the body temperature of 100.6 °F is considered unusual.
To find the percentage, we can use a standard normal distribution table or calculator to determine the area under the curve to the right of z = 3.87. This represents the proportion of individuals with temperatures higher than 100.6 °F.
The percentage can be calculated as (1 - probability) * 100. Since the probability is very close to 1, the percentage is very close to 0. Therefore, a very small percentage of normal and healthy adults would be considered to have a fever based on the cutoff of 100.6 °F.
This suggests that the cutoff of 100.6 °F may not be appropriate for considering a fever in healthy adults.