Answer :
The probability that a randomly selected adult has an IQ greater than 133.2 is 0.0436 or 4.36%.
To find the probability that a randomly selected adult has an IQ greater than 133.2, assuming adults have IQ scores that are normally distributed with a mean of 97.6 and a standard deviation of 20.9, follow these steps:
1. Calculate the z-score: z = (X - μ) / σ, where X is the IQ score, μ is the mean, and σ is the standard deviation.
z = (133.2 - 97.6) / 20.9
z ≈ 1.71
2. Use a z-table or a calculator to find the area to the left of the z-score, which represents the probability of having an IQ score lower than 133.2.
P(Z < 1.71) ≈ 0.9564
3. Since we want the probability of having an IQ greater than 133.2, subtract the area to the left of the z-score from 1.
P(Z > 1.71) = 1 - P(Z < 1.71) = 1 - 0.9564 = 0.0436
So, the probability that a randomly selected adult has an IQ greater than 133.2 is approximately 0.0436 or 4.36%.
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