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Here we are given that the mean of the IQ scores of adults is 100 and the standard deviation is 15. We need to find the probability that a randomly selected adult has an IQ greater than 131.5 .
Now, as we are given that the IQ values are normally distributed we use vaues from a normal distribution table.
Now z is equivalent to (variable - mean) / standard deviation. Or z= (131.5 - 100) / 15 = 2.1. From the normal distribution table the cumulative probability till z= 2.1 is 0.4821. As we need the probability of a person with an IQ greater than 131.5 it is equal to 0.5 - 0.4821 = 0.0179.
Therefore the required probability is 0.0179.
IQ x scores are normally distributed with mean , m = 100 and standard devition , s =15.
Therefore z = (x-m)/s is the standard normal variate.
Therefore when IQ , x = 131.5, z = (131.5-100)/15 = 2.1.
Therefore , the probability that x > 131.5 is as good as P(z > 2.1) .
P(z> 2.1) = 1- P(z<= 2.1), where P(z < = 2.1) could be obtained from the tables
P(z > = 2.1) = 1 - 0.98214 = 0.01786
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