Answer to Question #350136 in Statistics and Probability for ABC

Question

suppose that iq scores within a population are normally distributed with a mean 100 and a sd 10 area enclosed under normAL CURVE FOR IQ SCORES ABOVE 130 IS

Answer 1

We have the normal distribution with $\mu=100$ and $\sigma=10.$

Let's find the z-score for $X=130.$ :

$z=\frac{X-\mu}{\sigma}=\frac{130-100}{10}=3.$

The area enclosed under normal curve for iq scores above 130 equals the probability that X is greater than 130:

$P(X>130)=P(z>3)=1-(z<3)=1-0.9987=0.0013.$

Answer: 0.0013.