IQ scores are normally distributed with mean of 100 and a standard deviation of 15. If MENSA only accepts people with an IQ score at the 99″ percentile or higher; what is the lowest possible IQ score you can have and still be admitted t0 the organization? Mental disability has traditionally been diagnosed for anyone with an IQ of 70 or lower: By this standard, what percentage of the population would meet criteria to be diagnosed with mental disability? If a “normal IQ” is any score between 85 and 115, what proportion of the population has normal IQ score? What proportion of the population has an IQ score that is either lower than 92 or higher than 118?
The Correct Answer and Explanation is:
To answer these questions, we use the properties of the normal distribution, a bell-shaped curve that describes how a set of data is spread out. IQ scores are normally distributed with a mean (μ) of 100 and a standard deviation (σ) of 15.
1. MENSA Admission (99th percentile IQ):
To find the IQ score at the 99th percentile, we look for the z-score that corresponds to a cumulative probability of 0.9900.
From the z-table:
- z=2.33z = 2.33z=2.33 corresponds to the 99th percentile.
Now use the z-score formula to find the corresponding IQ score:X=μ+zσ=100+(2.33)(15)=100+34.95=134.95X = \mu + z\sigma = 100 + (2.33)(15) = 100 + 34.95 = 134.95X=μ+zσ=100+(2.33)(15)=100+34.95=134.95
Rounded to the nearest whole number, the lowest IQ for MENSA is:
👉 135
2. Mental Disability (IQ ≤ 70):
Calculate the z-score for an IQ of 70:z=X−μσ=70−10015=−2.00z = \frac{X – \mu}{\sigma} = \frac{70 – 100}{15} = -2.00z=σX−μ=1570−100=−2.00
Using the z-table:
- P(Z≤−2.00)=0.0228P(Z ≤ -2.00) = 0.0228P(Z≤−2.00)=0.0228
So, about 2.28% of the population has an IQ of 70 or lower.
👉 2.28% meet criteria for mental disability
3. Normal IQ Range (85 to 115):
Calculate z-scores:
- For 85: z=85−10015=−1.00z = \frac{85 – 100}{15} = -1.00z=1585−100=−1.00
- For 115: z=115−10015=1.00z = \frac{115 – 100}{15} = 1.00z=15115−100=1.00
From the z-table:
- P(Z≤1.00)=0.8413P(Z ≤ 1.00) = 0.8413P(Z≤1.00)=0.8413
- P(Z≤−1.00)=0.1587P(Z ≤ -1.00) = 0.1587P(Z≤−1.00)=0.1587
P(85≤IQ≤115)=0.8413−0.1587=0.6826P(85 ≤ IQ ≤ 115) = 0.8413 – 0.1587 = 0.6826P(85≤IQ≤115)=0.8413−0.1587=0.6826
So, 68.26% of the population has a normal IQ.
👉 68.26% have an IQ between 85 and 115
4. IQ < 92 or IQ > 118:
Find z-scores:
- For 92: z=92−10015=−0.53z = \frac{92 – 100}{15} = -0.53z=1592−100=−0.53
- For 118: z=118−10015=1.20z = \frac{118 – 100}{15} = 1.20z=15118−100=1.20
From the z-table:
- P(Z≤−0.53)=0.2981P(Z ≤ -0.53) = 0.2981P(Z≤−0.53)=0.2981
- P(Z≥1.20)=1−0.8849=0.1151P(Z ≥ 1.20) = 1 – 0.8849 = 0.1151P(Z≥1.20)=1−0.8849=0.1151
Add both probabilities:P(IQ<92 or IQ>118)=0.2981+0.1151=0.4132P(IQ < 92 \text{ or } IQ > 118) = 0.2981 + 0.1151 = 0.4132P(IQ<92 or IQ>118)=0.2981+0.1151=0.4132
👉 41.32% of the population falls outside the 92–118 range.
Summary of Results:
| Question | Answer |
|---|---|
| MENSA IQ (99th percentile) | 135 |
| % with IQ ≤ 70 | 2.28% |
| % with IQ 85–115 | 68.26% |
| % with IQ <92 or >118 | 41.32% |
This analysis uses standard normal distribution principles, converting raw IQ scores to z-scores, and then using the standard normal table (z-table) to find cumulative probabilities.
