Assume that adults have IQ scores that are normally distributed with a mean of 94 and a standard deviation of 14. Find the probability that a randomly selected adult has an IQ greater than 107.1. (Hint: Draw a graph.) To help visualize the area of interest, draw a standard normal curve. Label the given values for x and mu . x 94107.1

0.6517

Step-by-step explanation:

z = (x - μ)/σ

Z= standard score

x= observed value

μ= mean of the sample

σ= standard deviation of the sample

z = (x - μ)/σ  = (107.1 - 94 )/ 14 = 0.9357

probability that a randomly selected adult has an IQ greater than 107.1.  = P(Z > 0.935) =  0.6517

NB: the value is 0.6517 is pulled from the z table which can be found at the back of most math text.

i just took the test, the correct answer is .. 31.5 in.

answer: idk

step-by-step explanation: