Evaluate-32 + (2 – 6)(10) The answer is -49
1 answer:
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Answer:
IQ scores of at least 130.81 are identified with the upper 2%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 100 and a standard deviation of 15.
This means that
What IQ score is identified with the upper 2%?
IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.
IQ scores of at least 130.81 are identified with the upper 2%.