Assume the body temperature of healthy adults are normally distributed with a mean of 98.2F and a standard deviation of 0.62F.
If you have a body temperature of 99.00F, what is your percentile score?
Convert 99.00F to a standard score(or a z-score).
Is a body temprature of 99.00F unusual? Why or Why not?
Fifty adults are randomly selected. What is the likelyhood that the mean of their body temprature is 97.98F or lower?
A persons body temprature is found to be 101.00F. Is the result unusual? Why or Why not?
What body temprature is the 95th percentile?
What body temprature is in the 5th percentile?
Bellevue Hospital in NY uses 100.6F as the lowest temprature considered to indicate a fever. What percentage of normal and healthy adults would be considered to have a fever? Does this percentage suggust that a cutoff of 100.6F is appropriate?
To find the percentile score for a body temperature of 99.00F, we need to convert it to a standard score (z-score) using the formula:
z = (x - μ) / σ
where x is the value (99.00F), μ is the mean (98.2F), and σ is the standard deviation (0.62F):
z = (99.00 - 98.2) / 0.62
z = 0.806
To find the percentile score, we need to determine the percentage of values that fall below the z-score. We can use a standard normal distribution table or a calculator to find this value.
Looking up the z-score of 0.806 in a standard normal distribution table, we find that the percentile score is approximately 79.5%. This means that a body temperature of 99.00F is at the 79.5th percentile.
To determine if a body temperature of 99.00F is unusual, we need to compare it to the normal distribution. Unusual values are typically those that fall outside a certain range or percentile. In this case, a body temperature of 99.00F falls within the range of normal body temperatures, as it is above the mean and within the range of standard deviation. Therefore, it is not considered unusual.
To find the likelihood that the mean of 50 adults' body temperatures is 97.98F or lower, we need to calculate the z-score for this value:
z = (x - μ) / (σ / sqrt(n))
where x is the value (97.98F), μ is the mean (98.2F), σ is the standard deviation (0.62F), and n is the sample size (50).
z = (97.98 - 98.2) / (0.62 / sqrt(50))
z = -1.570
Using the standard normal distribution table or a calculator, we can determine that the likelihood of the mean body temperature being 97.98F or lower is approximately 6.71%.
A person's body temperature of 101.00F can be evaluated for "unusualness" by converting it to a z-score using the formula mentioned earlier.
z = (x - μ) / σ
z = (101.00 - 98.2) / 0.62
z = 4.52
Looking up the z-score of 4.52 in a standard normal distribution table or using a calculator, we find that the percentile score is almost 100%. This indicates that a body temperature of 101.00F is extremely rare within the normal distribution, making it unusual.
To find the body temperature corresponding to the 95th percentile, we need to find the z-score associated with the 95th percentile. The z-score for the 95th percentile is approximately 1.645. We can use this z-score to find the body temperature using the formula:
x = μ + (z * σ)
x = 98.2 + (1.645 * 0.62)
x ≈ 99.29F
Therefore, the body temperature at the 95th percentile is approximately 99.29F.
To find the body temperature corresponding to the 5th percentile, we need to find the z-score associated with the 5th percentile. The z-score for the 5th percentile is approximately -1.645. Using the same formula as above:
x = μ + (z * σ)
x = 98.2 + (-1.645 * 0.62)
x ≈ 96.83F
Therefore, the body temperature at the 5th percentile is approximately 96.83F.
To determine the percentage of normal and healthy adults considered to have a fever based on Bellevue Hospital's cutoff of 100.6F, we need to find the percentage of values that exceed this temperature. We can calculate the z-score for 100.6F using the formula mentioned earlier:
z = (x - μ) / σ
z = (100.6 - 98.2) / 0.62
z ≈ 3.87
From the z-score, we can determine the percentage of values greater than 100.6F using a standard normal distribution table or a calculator. The percentage is almost 0%, indicating that a very small proportion of normal and healthy adults would be considered to have a fever according to the cutoff of 100.6F.
This suggests that the cutoff of 100.6F may not be appropriate for determining if someone has a fever among normal and healthy adults, as it is likely to classify very few individuals as having a fever.