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Answer :
The probability that a randomly selected adult from this group has an IQ greater than 120.5 is approximately 0.4084.
To find the probability, we need to calculate the area under the normal distribution curve corresponding to IQ scores greater than 120.5. We can do this by standardizing the IQ score and using the standard normal distribution table or a statistical calculator.
Given:
Mean (μ) = 99.9
Standard deviation (σ) = 15.3
IQ score (X) = 120.5
First, we standardize the IQ score using the formula:
Z = (X - μ) / σ
Z = (120.5 - 99.9) / 15.3
≈ 1.343
Next, we find the area under the standard normal distribution curve to the right of Z = 1.343. Using a standard normal distribution table or a statistical calculator, we find the corresponding area to be approximately 0.0916.
However, we are interested in the probability of an IQ score greater than 120.5, which includes the area to the right of 120.5. Since the standard normal distribution is symmetric, we can subtract the area to the right of Z = 1.343 from 0.5 (which represents the area under the entire curve) to get the desired probability.
Probability = 0.5 - 0.0916
≈ 0.4084
Therefore, the probability that a randomly selected adult from this group has an IQ greater than 120.5 is approximately 0.4084.
The probability that a randomly selected adult from the given group has an IQ greater than 120.5 is approximately 0.4084.
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