Solve IQ Test Problem: Percent Adults Qualifying for MENSA Membership

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The discussion centers on calculating the percentage of adults qualifying for MENSA membership based on IQ scores, which follow a normal distribution with a mean of 100 and a standard deviation of 15. MENSA requires a minimum IQ score of 130 for membership. To determine the qualifying percentage, participants suggest using the 68-95-99.7 rule and standardizing the score using the formula Z = (X - μ) / σ. By applying this method, one can find the Z-score for 130 and subsequently use a standard normal distribution table to find the corresponding percentage. Understanding these calculations is essential for solving the problem effectively.
alandry06
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The problem goes as such;

The scale of scores on an IQ test is approximately Normal with mean 100 and standard deviation 15. The organization MENSA, which calls itself "the high-IQ society," requires an IQ score of 130 or higher for membership. What percent of adults would qualify for membership?


I know I can use the 68-95-99.7 rule, but I am confused at where to begin.
 
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Wrong section, sorry.
 
alandry06 said:
The problem goes as such;

The scale of scores on an IQ test is approximately Normal with mean 100 and standard deviation 15. The organization MENSA, which calls itself "the high-IQ society," requires an IQ score of 130 or higher for membership. What percent of adults would qualify for membership?


I know I can use the 68-95-99.7 rule, but I am confused at where to begin.
Can you state the the 68-95-99.7 rule?
 
You have to standardize first, using Z=X-\mu/\sigma
 
Lateral said:
You have to standardize first, using Z =
(X-\mu)/\sigma is what you must have meant.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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