IQ scores (as measured by the Stanford-Binet intelligence test) in a certain country are normally distributed with a mean of 95 and a standard deviation of 18. Find the approximate number of people in the country (assuming a total population of 323,000,000) with an IQ higher than 126. (Round your answer to the nearest hundred thousand.)

13,797,339 people

Step-by-step explanation:

Given that μ = 95 and σ = 18

The total number of population is N = 323,000,000

Now, we calculate the probability that IQ is higher than 126

P(X>126) = 1 - P[Z ≤ 126 - 95 / 18]

P(X>126) = 1 - P[Z ≤ 1.72]

P(X>126) = 1 - [ NORMSDIST(1.72)] Using MS Excel

P(X>126) = 1 - 0.957283779

P(X>126) = 0.042716221

Hence, the required probability is 0.042716221

Now, we calculate the approximate number of people in the county with an IQ higher than 126

Number of people = P(X>128) * N

Number of people = 0.042716221 * 323,000,000

Number of people = 13797339.383

Number of people = 13,797,339

Therefore, the approximate number of people in the country with an IQ higher than 126 is 13,797,339 in population

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)