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?

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

only 10/60 isn't equivalent

hope this : )

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: