Suppose IQ scores have a bell-shaped distribution with a mean of 100 and a standard deviation of 15. What percentage of the scores are above 130?
To find the percentage of scores that are above 130, we need to use the standard normal distribution table (also known as the Z-table) or a statistical calculator. I will explain both methods.
Method 1: Using a Z-table
1. Calculate the z-score associated with the IQ score of 130. The formula for calculating the z-score is: z = (x - μ) / σ, where x is the raw score, μ is the mean, and σ is the standard deviation.
z = (130 - 100) / 15
z = 2
2. Look up the area under the standard normal curve beyond z = 2 in the Z-table. The Z-table provides the cumulative probability (area) up to a given z-score. Since we want the area beyond 2, we need to subtract the cumulative probability from 1.
The cumulative probability for z = 2 is approximately 0.9772.
Area beyond z = 2 = 1 - 0.9772
Area beyond z = 2 = 0.0228
3. Convert the area to a percentage.
Percentage above 130 = Area beyond z = 2 * 100%
Percentage above 130 = 0.0228 * 100%
Percentage above 130 ≈ 2.28%
Method 2: Using a statistical calculator
You can also use a statistical calculator or software like Excel to find the percentage directly.
1. Use the cumulative distribution function (CDF) of the normal distribution with the given mean and standard deviation.
2. Calculate the CDF value for x = 130.
For example, in Excel, you can use the formula =NORM.DIST(130, 100, 15, TRUE) to calculate the CDF value for x = 130.
The result will be approximately 0.9772.
3. Subtract the CDF value from 1 and convert it to a percentage.
Percentage above 130 = (1 - CDF value) * 100%
Percentage above 130 = (1 - 0.9772) * 100%
Percentage above 130 ≈ 2.28%
Therefore, approximately 2.28% of the IQ scores are above 130.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score. Multiply by 100 to get percentage.