Question #350136

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


1
Expert's answer
2022-06-13T15:18:35-0400

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

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

z=Xμσ=13010010=3.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)=10.9987=0.0013.P(X>130)=P(z>3)=1-(z<3)=1-0.9987=0.0013.


Answer: 0.0013.


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