To find specific IQ scores based on a normally distributed population with a mean of 100 and a standard deviation of 15, you can use the properties of the normal distribution. However, since I cannot see the graph you referred to, I can provide you with a general method to find IQ scores based on various percentiles or probabilities.
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Identify the z-score: If you know the percentile, you can find the corresponding z-score using a z-table or a standard normal distribution calculator.
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Convert the z-score to an IQ score: The formula to convert a z-score to the corresponding IQ score is:
\[ X = \mu + (z \cdot \sigma) \]
where:
- \(X\) is the IQ score,
- \(\mu\) is the mean (100),
- \(z\) is the z-score,
- \(\sigma\) is the standard deviation (15).
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Example Calculations:
- For the 50th percentile (median), the z-score is 0. \[ X = 100 + (0 \cdot 15) = 100 \]
- For the 84th percentile (one standard deviation above the mean), the z-score is approximately 1. \[ X = 100 + (1 \cdot 15) = 115 \]
- For the 16th percentile (one standard deviation below the mean), the z-score is approximately -1. \[ X = 100 + (-1 \cdot 15) = 85 \]
If you have specific percentiles or probabilities in mind, please provide them, and I can help you determine the corresponding IQ scores using the methodology described!
2 answers