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

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.

, where

z = z-score,

x = Random sample score,

= Mean,

= Standard deviation.

Upon substituting our given values in z-score formula, we will get:

Now, we will use normal distribution table to find the .

Using formula , we will get:

Therefore, the probability that a randomly selected adult has an IQ greater than 115.9 is 0.14917 or approximately 14.92%.

i am not sure :

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