Some IQ scores are standardized with a mean of 100 and a standard deviation of 16. Using the 68-95-99.7 Rule, Determine: 1. What percent of people would have an IQ score of no more than 68. (2).What percent of people would have an IQ score of 100 or above.
1. 68 is 2 sigma from the mean. 95% are within that distance of the mean, and 5% are outside it. Half of those (2.5%) are below it on the low side (below 68).
2. 50%. The mean always divides the distribution in that ratio: half above and half below.
To determine the percentage of people with an IQ score of no more than 68, we need to find the area under the normal distribution curve to the left of 68.
1. Calculate the z-score:
The formula for the z-score is: z = (x - μ) / σ
Where x is the IQ score, μ is the mean, and σ is the standard deviation.
For an IQ score of 68,
z = (68 - 100) / 16
z = -2
2. Look up the z-score:
We need to find the area to the left of -2 on the standard normal distribution curve. A z-table or calculator can be used to find this value.
The area to the left of -2 is approximately 0.0228 or 2.28%.
Therefore, approximately 2.28% of people would have an IQ score of no more than 68.
To determine the percentage of people with an IQ score of 100 or above, we need to find the area under the normal distribution curve to the right of 100.
1. Calculate the z-score:
For an IQ score of 100,
z = (100 - 100) / 16
z = 0
2. Look up the z-score:
We need to find the area to the right of 0 on the standard normal distribution curve. A z-table or calculator can be used to find this value, or we can subtract the value we found in the previous step from 1.
The area to the right of 0 is 0.5 (or 50%).
Therefore, approximately 50% of people would have an IQ score of 100 or above.