Respuesta :
Answer:
A person must score 131 or above to qualify for Mensa.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 15
We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.02
[tex]P( X > x) = P( z > \displaystyle\frac{x - 100}{15})=0.02[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 100}{15})=0.02[/tex]
[tex]=P( z \leq \displaystyle\frac{x - 100}{15})=0.98[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 2.054) = 0.98[/tex]
[tex]\displaystyle\frac{x - 100}{15} = 2.054\\\\x = 130.81 \approx 131[/tex]
Thus, a person must score 131 or above to qualify for Mensa.
