Answer:
0.14917
Step-by-step explanation:
We have been given that adults have IQ scores that are normally distributed with a mean of 95.6 and a standard deviation of 19.5. We are asked to find the probability that a randomly selected adult has an IQ greater than 115.9.
First of all, we will find z-score corresponding to 115.9 using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where
z = z-score,
x = Random sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
Upon substituting our given values in z-score formula, we will get:
[tex]z=\frac{115.9-95.6}{19.5}[/tex]
[tex]z=\frac{20.3}{19.5}[/tex]
[tex]z=1.04[/tex]
Now, we will use normal distribution table to find the [tex]P(z>1.04)[/tex].
Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get:
[tex]P(z>1.04)=1-P(z<1.04)[/tex]
[tex]P(z>1.04)=1-0.85083[/tex]
[tex]P(z>1.04)=0.14917[/tex]
Therefore, the probability that a randomly selected adult has an IQ greater than 115.9 is 0.14917 or approximately 14.92%.