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Answer :
To calculate the probability of an adult having an IQ greater than 128.9, determine the z-score and use it to find the corresponding probability from the standard normal distribution. The z-score is 1.99, and the probability of a higher IQ score is 0.0233, which does not match any of the provided options.
The question asks us to calculate the probability that a randomly selected adult has an IQ greater than 128.9, given a normal distribution with a mean of 97.5 and a standard deviation of 15.8.
To find this probability, we first determine the z-score, which is the number of standard deviations a data point is from the mean. The formula for the z-score is:
Z = (X - \) / \
Where X is the value we are looking at (128.9 in this case), \ is the mean (97.5), and \ is the standard deviation (15.8). Plugging in the values, the z-score is:
Z = (128.9 - 97.5) / 15.8 = 1.99
The next step is to look up the z-score in a standard normal distribution table or use a calculator with normal distribution functions to find the probability that a z-score is greater than 1.99.
According to a standard normal distribution table, the area to the left of z=1.99 is approximately 0.9767. Since we are interested in the area to the right (the probability of having a higher score), we subtract this value from 1:
P(Z > 1.99) = 1 - P(Z < 1.99) = 1 - 0.9767 = 0.0233
Thus, the probability that a randomly selected adult has an IQ greater than 128.9 is approximately 0.0233, which is not one of the provided options.
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