To find the percentage of the population with IQ scores between 115 and 130 in a normally distributed population with a mean of 100 and a standard deviation of 15, we can use the standard normal distribution (Z-score).
First, we need to convert the IQ scores to Z-scores using the formula:
\[ Z = \frac{(X - \mu)}{\sigma} \]
where:
- \(X\) is the IQ score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
- Calculate the Z-score for 115:
\[ Z_{115} = \frac{(115 - 100)}{15} = \frac{15}{15} = 1 \]
- Calculate the Z-score for 130:
\[ Z_{130} = \frac{(130 - 100)}{15} = \frac{30}{15} = 2 \]
Next, we can use the standard normal distribution table or a calculator to find the area under the curve (the percentage of the population) for these Z-scores.
- For \(Z = 1\), the cumulative probability (area to the left) is approximately 0.8413.
- For \(Z = 2\), the cumulative probability is approximately 0.9772.
To find the percentage of the population with IQ scores between 115 and 130, we subtract the cumulative probabilities:
\[ P(115 < X < 130) = P(Z < 2) - P(Z < 1) = 0.9772 - 0.8413 = 0.1359 \]
Finally, convert this to a percentage:
\[ 0.1359 \times 100 \approx 13.6% \]
Therefore, the percentage of the population with IQ scores between 115 and 130 is approximately 13.6%.