The scale of scores on an IQ test is approximately Normal with mean 102 and standard deviation 18. The organization MENSA, which calls itself “the high IQ society,” requires an IQ score of 130 or higher for membership. Use this information to answer the following questions.
(a) What proportion of adults would qualify for membership?(Use 4 decimal places.)
(b) Corinne scored 118 on her IQ test. Her z-score is about(Use 2 decimal places.)
(c) Corinne scored 118 on her IQ test. She scored higher than what percent of all adults?% (Use 2 decimal places.)
The correct answer and explanation is :
To solve this problem, we will use the properties of the normal distribution. The IQ scores are normally distributed with a mean of 102 and a standard deviation of 18. For each part of the question, we will use the z-score formula and the standard normal distribution table (or a calculator) to get the required answers.
Part (a): Proportion of adults who would qualify for MENSA membership
MENSA requires an IQ score of 130 or higher. We want to find the proportion of adults who have an IQ score greater than or equal to 130.
Step 1: Find the z-score for an IQ score of 130.
The formula for calculating the z-score is:
[
z = \frac{X – \mu}{\sigma}
]
Where:
- (X) is the score (130),
- (\mu) is the mean (102),
- (\sigma) is the standard deviation (18).
[
z = \frac{130 – 102}{18} = \frac{28}{18} \approx 1.56
]
Step 2: Find the proportion to the right of this z-score.
We look up the z-score of 1.56 in the standard normal distribution table, or use a calculator. The area to the left of z = 1.56 is approximately 0.9418. Since we want the proportion of people with a score higher than 130, we subtract this value from 1:
[
P(X \geq 130) = 1 – 0.9418 = 0.0582
]
Thus, the proportion of adults who would qualify for MENSA membership is 0.0582 or 5.82%.
Part (b): Corinne’s z-score
Corinne scored 118 on her IQ test, and we need to calculate her z-score.
Step 1: Use the z-score formula.
[
z = \frac{118 – 102}{18} = \frac{16}{18} \approx 0.89
]
So, Corinne’s z-score is approximately 0.89.
Part (c): Corinne’s percentile
To find the percentile, we need to determine what percentage of adults scored lower than Corinne. This is equivalent to finding the cumulative area to the left of her z-score.
Step 1: Look up the area for z = 0.89.
The area to the left of z = 0.89 is approximately 0.8159. This means that Corinne scored higher than about 81.59% of all adults.
Thus, Corinne scored higher than 81.59% of adults.
Final Answers:
(a) Proportion qualifying for MENSA: 0.0582 (5.82%)
(b) Corinne’s z-score: 0.89
(c) Corinne’s percentile: 81.59%
Explanation:
- Z-scores: A z-score represents how many standard deviations a data point is from the mean. In this case, a score of 130 is 1.56 standard deviations above the mean of 102.
- Proportion for MENSA: The z-score of 1.56 corresponds to the top 5.82% of the population, which is why this is the proportion of people who would qualify for MENSA.
- Corinne’s percentile: A z-score of 0.89 means Corinne scored above 81.59% of all adults, indicating she is in the top 18.41% of the population.