Adults have IQ scores that are normally distibuted with a mean of 100 and a standard deviation of 15. Find the probability that a randomly selected adult has an IQ greater than 131.5.
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