Let X be a continuous random variable that is normally distributed with mean, µ = 15 and standard deviation, σ = 2.8, find a value xo that represents the 80th percentile of the distribution.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion, then insert the values into the equation above and solve for the score.
To find the value xo that represents the 80th percentile of the distribution, we can use the properties of the standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. In order to use the standard normal distribution to find percentiles of a normal distribution with a different mean and standard deviation, we need to convert our values into z-scores.
The z-score (also known as the standard score) represents the number of standard deviations an observation or value is from the mean. It is calculated using the formula:
z = (x - µ) / σ
Where:
- z is the z-score
- x is the value
- µ is the mean
- σ is the standard deviation
To find the value xo that represents the 80th percentile, we need to find the z-score corresponding to this percentile in the standard normal distribution.
The 80th percentile is equivalent to a percentile rank of 0.80. We can use a standard normal distribution table or a statistical software to find the z-score associated with the 80th percentile.
Using a standard normal distribution table, we can look up the area (probability) of 0.80 and find the corresponding z-score. From the table, we find that the z-score corresponding to a cumulative probability of 0.80 is approximately 0.84.
Now, we can go back to the formula for z-score and solve for x (the value we are interested in):
z = (x - µ) / σ
Rearranging the formula to solve for x:
x = z * σ + µ
Substituting the values we know:
x = 0.84 * 2.8 + 15
Calculating this expression:
x = 16.352
Therefore, the value xo that represents the 80th percentile of the distribution is approximately 16.352.