Assume that adults have IQ scores that are normally distributed with a mean of 95.695.6 and a standard deviation of 19.519.5. Find the probability that a randomly selected adult has an IQ greater than 115.9115.9

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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%.

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