Question
The IQ scores of a population are normally distributed with a mean of 100 and a standard deviation of 15. What percentage of the population would be recognized as having an IQ score between 115 and 130? Round to the nearest tenth.
A. 35.5%
B. 9.9%
C. 13.6%
D. 84.1%
A. 35.5%
B. 9.9%
C. 13.6%
D. 84.1%
Answers
GPT-4o mini
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 will use the properties of the normal distribution and the standard normal distribution (Z-scores).
1. **Calculate the Z-scores for 115 and 130**:
The Z-score formula is:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
- For 115:
\[
Z_{115} = \frac{(115 - 100)}{15} = \frac{15}{15} = 1
\]
- For 130:
\[
Z_{130} = \frac{(130 - 100)}{15} = \frac{30}{15} = 2
\]
2. **Find the percentages corresponding to these Z-scores using the standard normal distribution table**:
- The area to the left of \(Z = 1\) (which corresponds to an IQ of 115) is approximately 0.8413, or 84.13%.
- The area to the left of \(Z = 2\) (which corresponds to an IQ of 130) is approximately 0.9772, or 97.72%.
3. **Calculate the percentage of the population with an IQ between 115 and 130**:
To find the percentage between these two Z-scores, subtract the area at \(Z = 1\) from the area at \(Z = 2\):
\[
P(115 < X < 130) = P(Z < 2) - P(Z < 1) = 0.9772 - 0.8413 = 0.1359
\]
4. **Convert this to a percentage**:
\[
0.1359 \times 100 = 13.59\%
\]
When rounding to the nearest tenth, this result is approximately 13.6%.
Thus, the answer is **C. 13.6%**.
1. **Calculate the Z-scores for 115 and 130**:
The Z-score formula is:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
- For 115:
\[
Z_{115} = \frac{(115 - 100)}{15} = \frac{15}{15} = 1
\]
- For 130:
\[
Z_{130} = \frac{(130 - 100)}{15} = \frac{30}{15} = 2
\]
2. **Find the percentages corresponding to these Z-scores using the standard normal distribution table**:
- The area to the left of \(Z = 1\) (which corresponds to an IQ of 115) is approximately 0.8413, or 84.13%.
- The area to the left of \(Z = 2\) (which corresponds to an IQ of 130) is approximately 0.9772, or 97.72%.
3. **Calculate the percentage of the population with an IQ between 115 and 130**:
To find the percentage between these two Z-scores, subtract the area at \(Z = 1\) from the area at \(Z = 2\):
\[
P(115 < X < 130) = P(Z < 2) - P(Z < 1) = 0.9772 - 0.8413 = 0.1359
\]
4. **Convert this to a percentage**:
\[
0.1359 \times 100 = 13.59\%
\]
When rounding to the nearest tenth, this result is approximately 13.6%.
Thus, the answer is **C. 13.6%**.