Question

ch 3* the scores of adults on an iq test are 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?

o 5%
o 95%
o 2.5%

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 130$ (the IQ score for MENSA membership), $\mu = 100$ (mean), and $\sigma=15$ (standard deviation). So, $z=\frac{130 - 100}{15}=\frac{30}{15}=2$.

Step2: Use the standard normal distribution

We want to find $P(X\geq130)$, which is equivalent to $P(Z\geq2)$ in the standard - normal distribution. Since the total area under the standard - normal curve is 1, and the standard normal distribution is symmetric about $z = 0$, we know that $P(Z\geq2)=1 - P(Z < 2)$. Looking up $P(Z < 2)$ in the standard - normal table, we find that $P(Z < 2)=0.9772$. So, $P(Z\geq2)=1 - 0.9772 = 0.0228\approx2.5\%$.

Answer:

2.5%