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.