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You may need to use the appropriate appendix table or technology to answer this question. IQ scores (as measured by the Stanford-Binet intelligence test) in a certain country are normally distributed with a mean of 100 and a standard deviation of 19. Find the approximate number of people in the country (assuming a total population of 323,000,000) with an IQ higher than 128. (Round your answer to the nearest hundred thousand.)

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The first step in solving this problem is calculating the z-score. The z-score is a measure of how many standard deviations an element is from the mean. It is obtained by subtracting the mean IQ from the target IQ and then dividing the difference by the standard deviation. So, for an IQ of 128 in a population where the mean is 100 and standard deviation is 19, the z-score is (128 - 100) / 19, which yields approximately 1.47.

Then we want to find how much of the population has an IQ higher than the target, which is the same as finding the portion of the population that falls to the right of the z-score in a normal distribution. However, standard z-tables and functions often give the cumulative area to the left. So, to find the portion of the population to the right, we subtract the cumulative distribution function result from 1.

Taking the cumulative distribution function (CDF) for 1.47 results in approximately 0.93. Remember, this is the portion to the left. We need to subtract this from 1 to find the portion to the right: 1 - 0.93 which approximates to 0.07 or 7% of the population.

Finally, we want to predict the actual number of people in this country with an IQ over 128. The country's population is 323,000,000 people. So, multiply 0.07 (the portion of the population with an IQ over 128) by 323,000,000. This results in roughly 22,710,000. To round this to the nearest hundred thousand, we get 22,700,000.

Therefore, approximately 22,700,000 people in the nation have an IQ higher than 128.

Answer: There are approximately 22,700,000 people in the nation have an IQ higher than 128.

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