Assume the body temperature of healthy adults are normally distributed with a mean of 98.2 degrees F and a standard deviation of .62 degrees F. If you have a temperature of 99.0 what is your percentile score?
99 - 98.2 = .8
.8 divided by .6 = 1.29...z-score
z-score of 1.29...on table
=.9015 = 90.15 %
To determine the percentile score for a temperature of 99.0 degrees F in a normally distributed population, you can use the Z-score formula and the standard normal distribution table.
The Z-score formula calculates how far a given data point is from the mean of a distribution, in terms of standard deviations. The formula is:
Z = (X - μ) / σ
Where:
- Z is the Z-score,
- X is the given data point (99.0 degrees F in this case),
- μ is the mean of the distribution (98.2 degrees F), and
- σ is the standard deviation of the distribution (0.62 degrees F).
Plugging in the values, we get:
Z = (99.0 - 98.2) / 0.62
Z = 0.8 / 0.62
Z ≈ 1.29
Next, you need to find the percentile associated with the Z-score. You can do this by referring to a standard normal distribution table.
The standard normal distribution table provides you with the area under the curve for different Z-scores. This table typically gives you the values for the left tail of the curve (i.e., the area to the left of the Z-score).
Looking up the Z-score of 1.29 in the table, you find that the area to the left of this Z-score is approximately 0.9007.
To calculate the percentile score, subtract this area from 1 and convert it to a percentage:
Percentile = (1 - 0.9007) * 100
Percentile ≈ 9.93%
Therefore, if you have a temperature of 99.0 degrees F, your percentile score is approximately 9.93%.