Suppose that is normally distributed with mean 90 and standard deviation 12. What value of does only the top 20% exceed?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion and its Z score. Put Z and other values into the equation above and solve for the score.
To find the value of that only the top 20% exceed, we need to find the z-score associated with the top 20% of the normal distribution.
The z-score is a measure of how many standard deviations an observation is from the mean. It allows us to standardize a normal distribution and convert it into the standard normal distribution, which has a mean of 0 and a standard deviation of 1.
We can convert the problem into finding the z-score that corresponds to the top 20% of the normal distribution. Since the top 20% is the upper tail of the distribution, we need to find the z-score corresponding to a cumulative probability of 0.8.
To find this z-score, we can use a standard normal distribution table or a calculator. We can calculate the z-score using the following formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value of the variable
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In our case, we want to find x, so we need to rearrange the formula:
x = z * σ + μ
Now, we can substitute the given values into the equation:
μ = 90 (mean)
σ = 12 (standard deviation)
z = z-score corresponding to a cumulative probability of 0.8
Using the standard normal distribution table or a calculator, we can find that the z-score corresponding to a cumulative probability of 0.8 is approximately 0.84.
Plugging this into the equation, we can calculate the value of :
x = 0.84 * 12 + 90
x ≈ 101.08
Therefore, the value of that only the top 20% exceed is approximately 101.08.