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.
, 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%.
(a) 10 = 2x5
(b) 37= 1x37
(c) 25=5x5
Add 9 because -2 to 7 equals 9 hope this helped
Answer:
Step-by-step explanation: