suppose that iq scores have a bell-shaped distribution with a mean of 97 and a standard deviation of 17. using the empirical rule, what percentage of iq scores are greater than 46? please do not round your answer.
46 is 3 standard deviations below the mean
Here is the process.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability greater than Z.
To find the percentage of IQ scores that are greater than 46 using the empirical rule, we need to calculate the z-score for 46.
The z-score formula is: (X - μ) / σ, where X is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation.
Plugging in the values: X = 46, μ = 97, and σ = 17, we get:
z-score = (46 - 97) / 17
= -51 / 17
= -3
According to the empirical rule, approximately 99.7% of the data falls within 3 standard deviations of the mean. This means that the percentage of IQ scores greater than 46 would be the complement of the percentage within 3 standard deviations.
Since 99.7% of the data lies within 3 standard deviations, we can subtract this from 100% to find the percentage outside of that range.
Percentage outside of 3 standard deviations = 100% - 99.7%
= 0.3%
Therefore, the percentage of IQ scores greater than 46 is approximately 0.3%.
To find the percentage of IQ scores that are greater than a specific value using the empirical rule, we need to figure out how many standard deviations away from the mean that value falls.
The empirical rule states that in a bell-shaped distribution:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
In this case, we want to determine the percentage of IQ scores greater than 46.
First, we need to calculate how many standard deviations away from the mean 46 is.
Using the formula z = (x - μ) / σ, where:
- x is the value we want to find the number of standard deviations for (46 in this case),
- μ is the mean (97), and
- σ is the standard deviation (17),
we can substitute the values:
z = (46 - 97) / 17
z ≈ -3
Since 46 is 3 standard deviations below the mean, we know that approximately 99.7% of the data falls to the left of this value.
Therefore, the percentage of IQ scores greater than 46 is approximately 100 - 99.7 = 0.3%.