Answer: 0.0228 .
Step-by-step explanation:
Given, IQ scores are approximately normally distributed with a mean of 100 and standard deviation of 15
Let X denotes the IQ score.
Then, the proportion of people with IQs above 130 is
[tex]P(X>130)=P(\dfrac{X-mean}{ standard\ deviation}>\dfrac{130-100}{15})\\\\= P(Z>2)\ \ \ \[Z=\dfrac{X-mean}{ standard\ deviation}][/tex]
[tex]=1-P(Z<2)\ \ \ [P(Z>z)=1-P(Z<z)]\\\\=1-0.9772\ [\text{By z table}]\\\\=0.0228[/tex]
Hence, the proportion of people with IQs above 130 is 0.0228 .
