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The problem gives us that adults have an IQ score which is randomly distributed and has a mean of 100 and a standard deviation of 15. We need to find the probability that a randomly selected adult has an IQ between 110 and 120.
To do this we find the z-value which is given by z = (variable - mean)/standard deviation.
So for the IQ = 110, z = (110-100) / 15 = 2/3; and for IQ= 120, z = (120-100) / 15 = 4/3
Now we need to use a normal distribution table like the one provided here: Normal distribution table
For z = 4/3 we get the cumulative probability as 0.2454 and for z = 2/3 we get the cumulative probability as 0.4082. Therefore the probability between z=4/3 and z= 2/3 or the IQ between 120 and 110 is 0.4082 - 0.2454 = 0.1628
Therefore the required probability is 0.1628
Given that the IQ scores x are normally distributed with mean, m = 100 and standard deviation s = 15.
Therefore the z = (x-m)/s is a standard normal variate.
So to to find the probablity that the IQ lies between 110 and 120 is as good as the the standard normal variate z lies between (110-100)/15 = 2/3 and (120-100)/15 = 4/3.
Or P (2/3 < Z < 4/3) = P (z<= 4/3) - P(z < 2/3)
Or P(2/3 < z < 4/3) = P(z< 1.33)- P(z < = 0.67) = 0.90879-0.74751 = 0.16128. by standard normal tables.
P(2/3< z < 4/3) = 0. 0.16128.
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