Answer :
To solve the problem, we need to calculate two key values using the data provided: the standard deviation of the body temperatures and the probability of having a body temperature more than 97.8 degrees Fahrenheit using the Empirical Rule.
### Step 1: Standard Deviation
The standard deviation measures how spread out the numbers are in a dataset. Since we're dealing with a sample of adult body temperatures, we'll use the sample standard deviation formula.
1. Data: [tex]\(99.2, 98, 98.8, 98.3, 98.3, 99.7, 98.3, 98.9, 99.4, 98.7, 99.1\)[/tex]
2. Mean: The mean is given as 98.8.
3. Sample Standard Deviation Calculation:
- First, find the deviation of each temperature from the mean.
- Square each of these deviations.
- Sum all the squared deviations.
- Divide by the number of data points minus one (n-1, where n is the number of data points) to get the variance.
- Take the square root of the variance to get the standard deviation.
After performing these calculations, the standard deviation of the dataset is approximately 0.5 when rounded to the nearest tenth.
### Step 2: Probability of a Body Temperature More Than 97.8
We use the Empirical Rule, which applies to a normal distribution (bell curve), and z-scores to determine the probability.
1. Z-score Calculation:
- The z-score measures how many standard deviations an element is from the mean.
- Formula for z-score: [tex]\( z = \frac{{X - \text{{mean}}}}{{\text{{standard deviation}}}} \)[/tex]
- Here, [tex]\(X = 97.8\)[/tex], mean = 98.8, standard deviation = 0.5.
The calculated z-score is approximately [tex]\(-1.86\)[/tex].
2. Probability Calculation (using the z-score):
- The z-score table or a normal distribution calculator is used to find the probability of a z-score.
- The probability of getting a body temperature more than 97.8 is found by looking at the upper tail of the distribution.
Therefore, the probability of a body temperature being more than 97.8 degrees Fahrenheit is approximately 0.9688 when rounded to four decimal places.
In summary:
- The standard deviation of the body temperature data is approximately 0.5.
- The probability that a randomly selected adult has a body temperature more than 97.8 degrees Fahrenheit is approximately 0.9688.
### Step 1: Standard Deviation
The standard deviation measures how spread out the numbers are in a dataset. Since we're dealing with a sample of adult body temperatures, we'll use the sample standard deviation formula.
1. Data: [tex]\(99.2, 98, 98.8, 98.3, 98.3, 99.7, 98.3, 98.9, 99.4, 98.7, 99.1\)[/tex]
2. Mean: The mean is given as 98.8.
3. Sample Standard Deviation Calculation:
- First, find the deviation of each temperature from the mean.
- Square each of these deviations.
- Sum all the squared deviations.
- Divide by the number of data points minus one (n-1, where n is the number of data points) to get the variance.
- Take the square root of the variance to get the standard deviation.
After performing these calculations, the standard deviation of the dataset is approximately 0.5 when rounded to the nearest tenth.
### Step 2: Probability of a Body Temperature More Than 97.8
We use the Empirical Rule, which applies to a normal distribution (bell curve), and z-scores to determine the probability.
1. Z-score Calculation:
- The z-score measures how many standard deviations an element is from the mean.
- Formula for z-score: [tex]\( z = \frac{{X - \text{{mean}}}}{{\text{{standard deviation}}}} \)[/tex]
- Here, [tex]\(X = 97.8\)[/tex], mean = 98.8, standard deviation = 0.5.
The calculated z-score is approximately [tex]\(-1.86\)[/tex].
2. Probability Calculation (using the z-score):
- The z-score table or a normal distribution calculator is used to find the probability of a z-score.
- The probability of getting a body temperature more than 97.8 is found by looking at the upper tail of the distribution.
Therefore, the probability of a body temperature being more than 97.8 degrees Fahrenheit is approximately 0.9688 when rounded to four decimal places.
In summary:
- The standard deviation of the body temperature data is approximately 0.5.
- The probability that a randomly selected adult has a body temperature more than 97.8 degrees Fahrenheit is approximately 0.9688.