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).
Is a body temperature of 99.00 °F unusual?
Why or why not?
Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?
A person’s body temperature is found to be 101.00 °F.
Is the result unusual? Why or Why Not and What should you conclude?
What is your criterion for being "unusual"? Statistically the most often used criterion is P ≤ .05, meaning that the results would occur solely by chance only 5% of the time. P ≤ .01 is also used. P can be determined by first finding the Z score.
Z = (x - μ)/SD, where μ = mean
For the first question, Z = (99 - 98.2)/.62
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to this Z score.
Use the same method for the last temperature (101).
The Z score for the difference between means (second is a little different.
Z = (μ1 - μ2)/SE (Standard Error of the mean)
SE = SD/√(n-1), where √ = square root
Use the same table for that Z score.
You will have to reach your own conclusions. I hope this helps.
First part of the question: no, 99°F isn't all that unusual. It is (99.0 - 98.2) / 0.62 = 1.29 standard deviations above the mean - and if you look that up in a set of Normal distribution tables, you'll find that the area to the left of 1.29 is 0.9015. That means that 90.15% of the population have a body temperature of 99°F or less, so fractionally under 10% have a body temperature higher than that.
The standard error of the mean (SEM) of 50 patients is S/sqrt(N), where S is the standard deviation of the population, and N is the sample size. In our case that's 0.62/sqrt(50) = 0.088, so a mean of 97.98 is (98.20 - 97.98) / 0.088 = 2.5 SEMs beneath the population mean. Look up -2.5 (minus because it's to the left of the mean) in a set of Normal tables to get the area to the left of that value, and you'll get the answer to part 2. (Bear in mind that the area to the LEFT of -2.5 is the same as the area to the RIGHT of +2.5)
Finally, you should be able to tackle the third part of the question the same way as the first part.
To determine if a body temperature of 99.00 °F is unusual, we need to calculate its z-score.
1. Calculate the z-score:
Z = (X - μ) / σ
where
X = 99.00 °F (observed body temperature)
μ = 98.20 °F (mean body temperature)
σ = 0.62 °F (standard deviation)
Z = (99.00 - 98.20) / 0.62
Z = 0.80 / 0.62
Z ≈ 1.29
2. Check the z-score against the standard normal distribution table or calculator to determine if it is considered unusual.
From the standard normal distribution table, we can see that a z-score of 1.29 corresponds to a probability (or area under the curve) of approximately 0.9015. This means that 90.15% of the data falls below the body temperature of 99.00 °F.
Since 99.00 °F is within one standard deviation of the mean, with a probability of 0.9015, we can conclude that it is not unusual.
Now, let's calculate the likelihood that the mean of the body temperatures of 50 randomly selected adults is 97.98 °F or lower.
1. Calculate the standard error:
SE = σ / √n
where
σ = 0.62 °F (standard deviation of the adult body temperatures)
n = 50 (sample size)
SE = 0.62 / √50
SE ≈ 0.0876
2. Calculate the z-score for the mean temperature of 97.98 °F:
Z = (X - μ) / SE
where
X = 97.98 °F (mean body temperature)
μ = 98.20 °F (mean body temperature of the population)
Z = (97.98 - 98.20) / 0.0876
Z ≈ -0.22 / 0.0876
Z ≈ -2.512
3. Find the probability associated with this z-score using the standard normal distribution table or calculator.
From the standard normal distribution table, a z-score of -2.512 corresponds to a probability (or area under the curve) of approximately 0.0059. This means that the likelihood of randomly selecting 50 adults with a mean body temperature of 97.98 °F or lower is approximately 0.59% or 5.9 in 1000.
Therefore, the likelihood of randomly selecting 50 adults with a mean body temperature of 97.98 °F or lower is quite low.
Now, let's analyze a person’s body temperature of 101.00 °F to determine if it is unusual.
1. Calculate the z-score:
Z = (X - μ) / σ
where
X = 101.00 °F (observed body temperature)
μ = 98.20 °F (mean body temperature)
σ = 0.62 °F (standard deviation)
Z = (101.00 - 98.20) / 0.62
Z ≈ 2.26 / 0.62
Z ≈ 3.65
2. Check the z-score against the standard normal distribution table or calculator to determine if it is considered unusual.
From the standard normal distribution table, a z-score of 3.65 corresponds to a probability (or area under the curve) of approximately 0.9999. This means that 99.99% of the data falls below the body temperature of 101.00 °F.
Since 101.00 °F is more than three standard deviations away from the mean, with a probability of 0.9999, we can conclude that it is highly unusual. This could suggest a potential fever or other medical condition.
In conclusion:
- A body temperature of 99.00 °F is not unusual.
- The likelihood of a mean body temperature of 97.98 °F or lower among 50 randomly selected adults is approximately 0.59%.
- A person's body temperature of 101.00 °F is highly unusual and may indicate a potential fever or medical condition.
To determine whether a body temperature of 99.00 °F is unusual, we can use the concept of z-scores. A z-score measures how many standard deviations a value is away from the mean.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x is the given value (99.00 °F in this case)
μ is the mean (98.20 °F)
σ is the standard deviation (0.62 °F)
Calculating the z-score:
z = (99.00 - 98.20) / 0.62 = 1.29
Now, we refer to a Z-table or use statistical software to find the corresponding area/probability associated with the z-score. This will tell us how likely it is for a value to occur given the distribution.
From the Z-table, we find that a z-score of 1.29 corresponds to a cumulative probability of approximately 0.9015. This means that 90.15% of the data falls below a body temperature of 99.00 °F.
Since the probability is relatively high (more than 5%), we can conclude that a body temperature of 99.00 °F is not considered unusual. It falls within a normal range of body temperatures for healthy adults.
Now, let's move on to the second part of the question.
To find the likelihood that the mean of the body temperatures of fifty randomly selected adults is 97.98 °F or lower, we can use the Central Limit Theorem. The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases.
Since we have a sample size of 50, we can assume that the distribution of the sample means will follow a normal distribution. The mean of the sample means will be the same as the population mean, which is 98.20 °F. However, the standard deviation of the sample means, also known as the standard error, can be calculated using the formula σ / √n, where n is the sample size.
Using the given standard deviation of 0.62 °F and the sample size of 50, we can calculate the standard error:
standard error = 0.62 / √50 ≈ 0.0877
Now, we want to find the probability that the sample mean is 97.98 °F or lower. So, we can calculate the z-score using the formula:
z = (x - μ) / standard error
Calculating the z-score:
z = (97.98 - 98.20) / 0.0877 ≈ -0.251
From the Z-table or statistical software, we find that a z-score of -0.251 corresponds to a cumulative probability of approximately 0.4013. This means that there is a 40.13% chance that the mean body temperature of fifty randomly selected adults is 97.98 °F or lower.
Finally, let's address the last part of the question.
A person's body temperature is found to be 101.00 °F. To determine if this result is unusual, we can again use the z-score formula. However, this time we'll use the individual's body temperature in relation to the mean and standard deviation of the population.
Calculating the z-score:
z = (101.00 - 98.20) / 0.62 ≈ 4.52
From the Z-table or statistical software, we find that a z-score of 4.52 corresponds to a cumulative probability very close to 1. This means that a body temperature of 101.00 °F is extremely unlikely to occur within a population of healthy adults with a normal distribution of body temperatures.
Therefore, we can conclude that a body temperature of 101.00 °F is considered unusual, given the mean and standard deviation provided. It may indicate a potential abnormality or illness, and further medical attention may be necessary.