
Russ P. answered 01/04/15
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Nora,
IQ is a random variable which is approximately Normally distributed as N(μ,σ) = N(100, 15). Since IQ cannot be negative or very large positive (I think the highest ever recorded is somewhere around 280), it is not exactly Normally distributed since the Normal function has tails on both sides that go to infinity. However, within the practical "middle" range of us mere mortals it provides a very good approximation so we use it.
The Normal distribution is is a symmetrical distribution about its mean. Usually tables of values are only given for the right half so if you have a "less than" question, you must add a probability of 0.5 to the tabular data. Moreover, the data are only supplied for a normalized or standardized version N(0,1) of it, so you first have to transform your random variable x to an "equivalent" standardized z-value as follows:
z = (x - μ)/σ . In your case x = IQ. With that preamble, let's answer your 2 questions.
(1) % of population with IQ < 131?
z1 = (131 - 100)/15 = 31/15 = 2.067
Probability that z1 < 2.067 = 0.50 (adj for z1 <0) + Prob(0 ≤ z1 < 2.067 [from N(0,1) area table]
= 0.50 + 0.4806 (interpolated)
= 0.98 (rounded off)
So 98 % of the general population have IQs below 131, and only about 2% exceed that number.
(2) % of population with IQ < 105?
z2 = (105 - 100)/15 = 5/15 = 0.333
Probability that z2 < 0.333 = 0.50 (adj for z2 <0) + Prob(0 ≤ z2 < 0.333 [from N(0,1) area table]
= 0.50 + 0.1306 (interpolated)
= 0.63 (rounded off)
So 63 % of the general population have IQs below 105, and only about 37% exceed that number.
z2 = (105 - 100)/15 = 5/15 = 0.333
Probability that z2 < 0.333 = 0.50 (adj for z2 <0) + Prob(0 ≤ z2 < 0.333 [from N(0,1) area table]
= 0.50 + 0.1306 (interpolated)
= 0.63 (rounded off)
So 63 % of the general population have IQs below 105, and only about 37% exceed that number.