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%.
The formula to find<span> a </span>circle's area<span> (radius)</span>2<span> usually expressed as π ⋅ r 2 where r is the radius of a </span>circle<span>. </span>Area<span> of </span>Circle<span> Concept. The </span>area of a circle<span> is all the space inside a </span>circle's<span> circumference.</span>
The answer is A. (2,2)....
Answer:
median 1B: 310
Step-by-step explanation:
For the second question (5/6)/(4) is the same as (5/6)*(1/4) so just multiply across and you get (5/24)