Human iq scores are approximately normally distributed with mean 100 and standard deviation 15. determine the minimum iq scores for the top 5% of the population.

To solve this problem, we make use of the z statistic. A population of 5% means that we are looking for the population at >95%, P = 0.95. Using the standard distribution tables for z, a value of P = 0.95 indicates a value of z of z = 1.645

Now given the z and standard deviation s and the mean u, we can calculate for the value of IQ of the top 5% (x):

x = z s + u

x = 1.645 (15) + 100

x = 24.675 + 100

x = 124.675

Therefore the minimum iq score for the top 5% of the population is around 124.675

step-by-step explanation:

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.