Question

Suppose that IQ scores within a population are normally distributed with mean 100 and standard deviation 10. If we pick a person at random from this population, what it the probability that he or she has an IQ score

(a) above 130?
(b) below 110?
(c) between 90 and 120?

Answer 1

Answer:

a) Probability is 0.001

b) Below 110 is 0.84

c) Between 90 and 120 is 0.818

Step-by-step explanation:

We can calculate this using the cumulative normal distribution , by first standarizing our random variable X defined as IQ Scores  to a Z standard normal variable ( mean 1 - std 0 ) for the specific value we need :

a)

Z=X-mean/std = 130-100/10 =3.00

Then we can calculate:

P(X>130) = P(Z>3) = 1- P(Z<3) due to rules of probability.

This probability can be calculated with the function Norm.Distr(130,100,10,True) in  Excel or using the Z table looking for the value of Z = 3 in the rows and the value of the hundredths in Z in this case 0.00. If we do it in  excel we get:

1-P(Z<3)=1-0.9986=0.0013

b) We execute the same procedure as above now for 110

Z=110-100/10=1

P(X<110)=P(Z<1)=0.84

c) In this case the interval can be calculated as the area between the large an the low values.

Z1=120-100/10 = 2

Z2 =90-100/10= -1

P(90<X<120) = P(X<120)-P(X<90)=P(Z1<2)-P(Z2<-1)=0.977-0.158=0.818