Answer :
Final answer:
The probability that a randomly selected adult would have an IQ score greater than 123.4 is approximately 0.132 or 13.2%, based upon a normal distribution with a mean of 97.5 and standard deviation of 23.1.
Explanation:
To start, we need to standardize the IQ score by using the z-score formula, which is defined as Z = (X - μ) / σ. Where X is the value we are interested in (in this case the IQ score of 123.4), μ is the mean of the data set (97.5), and σ is the standard deviation (23.1).
Our z-value would therefore be: Z = (123.4 - 97.5) / 23.1 = 1.12. Now, having the z-score, we can use a z-table or any statistical software to find its corresponding probability, which is the probability that a randomly selected adult has an IQ lower than 123.4.
The value from the z-table for 1.12 is approximately 0.868, meaning 86.8% of the population has an IQ score less than 123.4. Therefore, the probability that a randomly selected adult has an IQ scores greater than 123.4 is 1 - 0.868 = 0.132. Thus, there is a 13.2% chance that a randomly selected adult will have an IQ score of over 123.4.
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