Answer: 0.1335
Step-by-step explanation:
Given: Children's IQs follow a normal distribution with mean [tex]\mu=100[/tex] and standard deviation of [tex]\sigma=9[/tex].
Let X be the IQ of a random child.
The probability that a randomly selected child has IQ above 110 = P(X>100)
[tex]=P(\dfrac{X-\mu}{\sigma}>\dfrac{110-100}{9})\\\\=P(Z>1.11)\\\\=1-P(Z<1.11)\ \ \ \ [P(Z>z)=1-P(Z<z)]\\\\=1-0.8665\\\\=0.1335\ \ \ [\text{By p-value table}][/tex]
Hence, the probability that a randomly selected child has IQ above 110= 0.1335
