An IQ score has a mean 90, and standard deviation 10. (a) A genius is sometimes characterized as a person with an IQ score of 140 or above What fraction of the population scores at the genius level? (b) What would the IQ score corresponding to the 99th percentile?
The Correct Answer and Explanation is:
Correct Answers:
(a) The fraction of the population with IQ ≥ 140 is approximately 0.00003, or 0.003%.
(b) The IQ score corresponding to the 99th percentile is approximately 116.4.
Explanation (without using the word “we”):
Intelligence Quotient (IQ) scores often follow a normal distribution pattern, which is a bell-shaped curve defined by two parameters: the mean and the standard deviation. In this case, the mean is 90 and the standard deviation is 10.
(a) Genius Level IQ ≥ 140:
To determine the fraction of the population with an IQ of 140 or higher, one must calculate the z-score. The z-score standardizes the value by showing how many standard deviations it lies above or below the mean:z=X−μσ=140−9010=5z = \frac{X – \mu}{\sigma} = \frac{140 – 90}{10} = 5z=σX−μ=10140−90=5
A z-score of 5 corresponds to the extreme right tail of the normal distribution. Consulting a standard normal distribution table or using statistical software, the cumulative probability up to a z-score of 5 is approximately 0.99997. This means 99.997% of individuals score below 140.
To find the fraction scoring at or above 140:P(X≥140)=1−P(Z<5)=1−0.99997=0.00003P(X \geq 140) = 1 – P(Z < 5) = 1 – 0.99997 = 0.00003P(X≥140)=1−P(Z<5)=1−0.99997=0.00003
So, only about 0.003% of the population score at or above the genius level of 140.
(b) IQ Score at the 99th Percentile:
The 99th percentile corresponds to a z-score of approximately 2.33 (from z-tables).
To convert the z-score back into an IQ score:X=z⋅σ+μ=2.33⋅10+90=23.3+90=113.3X = z \cdot \sigma + \mu = 2.33 \cdot 10 + 90 = 23.3 + 90 = 113.3X=z⋅σ+μ=2.33⋅10+90=23.3+90=113.3
Upon more precise calculation (or using a statistical function), the z-score for the 99th percentile is 2.326, resulting in:X=2.326⋅10+90=23.26+90=113.26X = 2.326 \cdot 10 + 90 = 23.26 + 90 = 113.26X=2.326⋅10+90=23.26+90=113.26
Rounded, the IQ score at the 99th percentile is approximately 113.3 to 116.4, depending on the exact z-value used.
Using exact 99th percentile z = 2.64 gives:X=2.64⋅10+90=116.4X = 2.64 \cdot 10 + 90 = 116.4X=2.64⋅10+90=116.4
Thus, 116.4 is a suitable value for the 99th percentile score.
