Assume the body temperature of healthy adults are normally distributed with a mean of 98.20 degrees F (based on data from the Uniservity of Maryland researchers).a. If you have a body temperature of 99.oo 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 liklihood that the meand of their body temperatures is 97.98 degrees F or lower? e. A person's body temperature is found to be 101.00 degrees f. Is the result unusual? why or why not? f. What body temoerature is the 95th percentile? g. What body temperature is the 5th percentile? h. Bellevue Hospital in New York uses 100.6 degrees 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 degrees F is appropriate?
What is the standard deviation?
a, b. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
c. Same as above.
d. Z = (score-mean)/SEm
SEm = SD/√n
Use same table.
e. Same process as a.
f, g. Start with the table and insert the Z score in the top equation.
h. Same process as a.
a. To find the percentile score for a body temperature of 99.00 degrees F, we need to calculate the area under the curve to the left of 99.00 degrees F in the normal distribution.
To do this, we can use a statistical calculator or a standard normal distribution table. Since we have the mean and standard deviation are not provided, we'll assume a standard deviation of 0.75 based on typical body temperature variability.
Using a standard normal distribution table, we can find the percentile corresponding to a z-score of (99.00 - 98.20)/0.75 = 1.07.
The percentile score can be determined by looking up the z-score in the table, which gives us a value of approximately 0.8577. This means that a body temperature of 99.00 degrees F corresponds to the 85.77th percentile.
b. To convert 99.00 degrees F to a standard score or z-score, we can use the formula:
z = (x - mean) / standard deviation
Using the mean of 98.20 degrees F and assuming a standard deviation of 0.75, the z-score would be:
z = (99.00 - 98.20) / 0.75 = 1.07
So, the z-score for a body temperature of 99.00 degrees F is 1.07.
c. To determine if a body temperature of 99.00 degrees F is unusual, we can compare its z-score to a standard normal distribution. Unusual values typically fall outside of the range of approximately ±2 standard deviations from the mean. However, it's important to note that the definition of what is considered unusual can vary depending on the context.
Since we assumed a standard deviation of 0.75, a body temperature of 99.00 degrees F has a z-score of 1.07, which falls within the range of ±2 standard deviations. Therefore, it is not considered unusually high.
d. To find the likelihood that the mean of fifty adults' body temperatures is 97.98 degrees F or lower, we need to use the sampling distribution of the mean. Assuming the body temperatures are normally distributed, the distribution of the sample means will also be normally distributed.
The standard deviation of the sample mean is given by the population standard deviation divided by the square root of the sample size. In this case, since the population standard deviation is not provided, we'll assume 0.75.
The standard deviation of the sample mean is: 0.75 / sqrt(50) = 0.106
To find the likelihood, we need to calculate the z-score:
z = (97.98 - 98.20) / 0.106 ≈ -2.07
Using a standard normal distribution table or a statistical calculator, we can find the likelihood associated with a z-score of -2.07. The area to the left of -2.07 is approximately 0.0188, or 1.88%.
Therefore, the likelihood that the mean of fifty adults' body temperatures is 97.98 degrees F or lower is approximately 1.88%.
e. To determine if a body temperature of 101.00 degrees F is unusual, we can follow a similar approach as before and compare the z-score to a standard normal distribution.
Using the assumed mean of 98.20 degrees F and standard deviation of 0.75, the z-score for a body temperature of 101.00 degrees F can be calculated as:
z = (101.00 - 98.20) / 0.75 ≈ 3.73
A z-score of 3.73 is well beyond ±2 standard deviations from the mean. Therefore, a body temperature of 101.00 degrees F would be considered unusual.
f. To find the body temperature corresponding to the 95th percentile, we can use a standard normal distribution table or a statistical calculator.
The 95th percentile corresponds to a z-score of approximately 1.645.
Using the formula:
x = (z * standard deviation) + mean
With the assumed mean of 98.20 degrees F and standard deviation of 0.75, the temperature at the 95th percentile is approximately:
x = (1.645 * 0.75) + 98.20 ≈ 99.88 degrees F
Therefore, the body temperature at the 95th percentile is approximately 99.88 degrees F.
g. Similarly, to find the body temperature corresponding to the 5th percentile, we can use a standard normal distribution table or a statistical calculator.
The 5th percentile corresponds to a z-score of approximately -1.645.
Using the formula:
x = (z * standard deviation) + mean
With the assumed mean of 98.20 degrees F and standard deviation of 0.75, the temperature at the 5th percentile is approximately:
x = (-1.645 * 0.75) + 98.20 ≈ 96.87 degrees F
Therefore, the body temperature at the 5th percentile is approximately 96.87 degrees F.
h. To determine the percentage of normal and healthy adults considered to have a fever according to Bellevue Hospital's cutoff of 100.6 degrees F, we can use the concept of z-scores and the standard normal distribution.
First, we need to calculate the z-score corresponding to a temperature of 100.6 degrees F using the mean and standard deviation provided.
z = (100.6 - 98.20) / (0.75) ≈ 3.20
Using a standard normal distribution table or a statistical calculator, we can find the percentage of observations beyond the z-score of 3.20.
Typically, we would look for the percentage in the tail area to the right of the z-score. However, in this case, we are interested in the percentage that falls below the cutoff temperature, so we need to consider the tail area to the left of the z-score.
The tail area to the left of 3.20 is very close to zero, which means an extremely small percentage of normal and healthy adults would be considered to have a fever according to Bellevue Hospital's cutoff of 100.6 degrees F.
This suggests that using a cutoff of 100.6 degrees F may not be appropriate for identifying fever in normal and healthy adults, as it would classify a very small percentage as having a fever.