Answer :
Final answer:
The probability that a randomly chosen adult has an IQ score greater than 136.6 is approximately 2.56%, obtained by calculating a z-score and using a z-table to find the cumulative probability.
Explanation:
To solve this problem, we'll use the concept of z-scores in statistics. A z-score tells us how many standard deviations away a certain value is from the mean.
Calculate the z-score for an IQ of 136.6: z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. So, z = (136.6 - 101.2) / 18.2 = 1.947.
Next, we consult a standardized z-table, which shows the cumulative probabilities for a standard normal distribution (mean of 0 and standard deviation of 1).
For a z-score of 1.947, the cumulative probability is approximately 0.9744. This means there's a 97.44% probability that an individual selected at random will have a score of 136.6 or less.
However, the question asks for the probability that an IQ score will be greater than 136.6. To find this, we subtract the cumulative probability from 1. So, the required probability = 1 - 0.9744 = 0.0256 or 2.56% when rounded to four decimal places.
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