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. 15.9%
B. 35.5%
C. 9.9%
D. 84.1% 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. 15.9%
B. 35.5%
C. 9.9%
D. 84.1%
A. 15.9%
B. 35.5%
C. 9.9%
D. 84.1% 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. 15.9%
B. 35.5%
C. 9.9%
D. 84.1%
Answers
GPT-4o mini
To find the percentage of the population with IQ scores between 115 and 130, we can use the properties of the normal distribution.
1. **Standardize the scores (z-scores)**:
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 \(X = 115\):
\[
z_{115} = \frac{(115 - 100)}{15} = \frac{15}{15} = 1
\]
- For \(X = 130\):
\[
z_{130} = \frac{(130 - 100)}{15} = \frac{30}{15} = 2
\]
2. **Find the area under the normal curve**:
We will look up the z-scores in the standard normal distribution table, or use a calculator.
- The cumulative probability for \(z = 1\) is approximately \(0.8413\) (or 84.13%).
- The cumulative probability for \(z = 2\) is approximately \(0.9772\) (or 97.72%).
3. **Calculate the percentage between the two z-scores**:
To find the percentage of the population that falls between these z-scores, subtract the cumulative probability of \(z = 1\) from the cumulative probability of \(z = 2\):
\[
P(115 < X < 130) = P(Z < 2) - P(Z < 1) = 0.9772 - 0.8413 = 0.1359
\]
Converting this to a percentage:
\[
0.1359 \times 100 \approx 13.59\%
\]
Given the options, the closest to this value is:
**B. 35.5%**.
(Note: There appears to be a mistake since 13.59% does not match any given option. Correctly interpreting the ranges and checking values helps ensure accurate results.)
1. **Standardize the scores (z-scores)**:
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 \(X = 115\):
\[
z_{115} = \frac{(115 - 100)}{15} = \frac{15}{15} = 1
\]
- For \(X = 130\):
\[
z_{130} = \frac{(130 - 100)}{15} = \frac{30}{15} = 2
\]
2. **Find the area under the normal curve**:
We will look up the z-scores in the standard normal distribution table, or use a calculator.
- The cumulative probability for \(z = 1\) is approximately \(0.8413\) (or 84.13%).
- The cumulative probability for \(z = 2\) is approximately \(0.9772\) (or 97.72%).
3. **Calculate the percentage between the two z-scores**:
To find the percentage of the population that falls between these z-scores, subtract the cumulative probability of \(z = 1\) from the cumulative probability of \(z = 2\):
\[
P(115 < X < 130) = P(Z < 2) - P(Z < 1) = 0.9772 - 0.8413 = 0.1359
\]
Converting this to a percentage:
\[
0.1359 \times 100 \approx 13.59\%
\]
Given the options, the closest to this value is:
**B. 35.5%**.
(Note: There appears to be a mistake since 13.59% does not match any given option. Correctly interpreting the ranges and checking values helps ensure accurate results.)