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Answer :
The probability that a randomly selected adult has an IQ greater than 126.9 can be found by calculating the z-score and using the standard normal distribution table.
To find the probability, we first calculate the z-score using the formula: z = (x - μ) / σ, where x is the IQ score, μ is the mean, and σ is the standard deviation.
Plugging in the values, we have z = (126.9 - 98.3) / 24.9 = 1.148.
Next, we use the standard normal distribution table (z-table) to find the probability corresponding to a z-score of 1.148. Looking up this value in the table, we find that the probability is approximately 0.8749.
Therefore, the probability that a randomly selected adult has an IQ greater than 126.9 is approximately 0.8749, or 87.49%.
Learn more about z-scores here: brainly.com/question/31871890
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