IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose an individual is randomly chosen. a) Find the probability that the person has an IQ greater than 125. b) Find the probability that the person has an IQ score between 105 and 118. c) What is the IQ score of a person whose percentile rank is at the 75th percentile, P75? d) Use the information from part (c) to fill in the blanks and circle the correct choice in the following statement. % of the individuals (persons) have an IQ score less than/more than __ e) “MENSA” is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the “MENSA” organization.
The Correct Answer and Explanation is:
To solve the questions, we use the normal distribution, which is symmetric and bell-shaped. Given:
- Mean (μ) = 100
- Standard deviation (σ) = 15
Let IQ be represented by the variable X∼N(100,152)X \sim N(100, 15^2)X∼N(100,152).
a) P(X > 125)
Step 1: Convert to z-scorez=X−μσ=125−10015=2515≈1.67z = \frac{X – \mu}{\sigma} = \frac{125 – 100}{15} = \frac{25}{15} \approx 1.67z=σX−μ=15125−100=1525≈1.67
Step 2: Find P(Z > 1.67)
Using the standard normal distribution table:P(Z>1.67)=1−P(Z≤1.67)=1−0.9525=0.0475P(Z > 1.67) = 1 – P(Z \leq 1.67) = 1 – 0.9525 = 0.0475P(Z>1.67)=1−P(Z≤1.67)=1−0.9525=0.0475
Answer:P(X>125)=0.0475P(X > 125) = 0.0475P(X>125)=0.0475
b) P(105 < X < 118)
Step 1: Convert both to z-scoresz1=105−10015=515=0.33z_1 = \frac{105 – 100}{15} = \frac{5}{15} = 0.33z1=15105−100=155=0.33z2=118−10015=1815=1.20z_2 = \frac{118 – 100}{15} = \frac{18}{15} = 1.20z2=15118−100=1518=1.20
Step 2: Find the probabilitiesP(0.33<Z<1.20)=P(Z<1.20)−P(Z<0.33)P(0.33 < Z < 1.20) = P(Z < 1.20) – P(Z < 0.33)P(0.33<Z<1.20)=P(Z<1.20)−P(Z<0.33)
From the z-table:P(Z<1.20)=0.8849,P(Z<0.33)=0.6293P(Z < 1.20) = 0.8849,\quad P(Z < 0.33) = 0.6293P(Z<1.20)=0.8849,P(Z<0.33)=0.6293P(105<X<118)=0.8849−0.6293=0.2556P(105 < X < 118) = 0.8849 – 0.6293 = 0.2556P(105<X<118)=0.8849−0.6293=0.2556
Answer:P(105<X<118)=0.2556P(105 < X < 118) = 0.2556P(105<X<118)=0.2556
c) Find P₇₅ (75th percentile)
Use z-tables to find the z-score corresponding to the 75th percentile:P(Z<z)=0.75⇒z≈0.674P(Z < z) = 0.75 \Rightarrow z \approx 0.674P(Z<z)=0.75⇒z≈0.674
Convert z to IQ score:X=μ+zσ=100+(0.674)(15)≈100+10.11=110.11X = \mu + z\sigma = 100 + (0.674)(15) \approx 100 + 10.11 = 110.11X=μ+zσ=100+(0.674)(15)≈100+10.11=110.11
Answer:P75=110.11P_{75} = 110.11P75=110.11
d) Interpret the percentile
From part (c):
“75% of individuals have an IQ score less than 110.11.”
Answer:75% of the individuals have an IQ score less than 110.11\boxed{75\%} \text{ of the individuals have an IQ score } \boxed{\text{less than 110.11}}75% of the individuals have an IQ score less than 110.11
e) Minimum IQ for MENSA (Top 2%)
Top 2% → 98th percentile
Find the z-score for the 98th percentile:P(Z<z)=0.98⇒z≈2.054P(Z < z) = 0.98 \Rightarrow z \approx 2.054P(Z<z)=0.98⇒z≈2.054
Convert to IQ score:X=μ+zσ=100+(2.054)(15)≈100+30.81=130.81X = \mu + z\sigma = 100 + (2.054)(15) \approx 100 + 30.81 = 130.81X=μ+zσ=100+(2.054)(15)≈100+30.81=130.81
Answer:Minimum IQ to qualify for MENSA=130.81\text{Minimum IQ to qualify for MENSA} = 130.81Minimum IQ to qualify for MENSA=130.81
Summary
| Part | Result |
|---|---|
| a) | P(X > 125) = 0.0475 |
| b) | P(105 < X < 118) = 0.2556 |
| c) | 75th percentile = 110.11 |
| d) | 75% have IQ less than 110.11 |
| e) | Minimum IQ for MENSA = 130.81 |
These calculations demonstrate how the normal distribution is used to model real-world data such as intelligence scores, allowing for meaningful interpretation of probabilities and percentiles.
