Question

Consider a population with mean 100 and standard deviation of 10. If we randomly select a data point from the population, what is the probability that the point will be between 90 and 110? (Hint: consider the Z-scores)?
What percentage of the population lies between 900 and 1100?                  

Answer 1

Solution:

What is the probability that the point will be between 90 and 110?

We are given that the population have mean = 100 and SD = 10.

We have to find P(90<X<110)

P(90<X<110) = P(X<110) – P(X<90)

Find P(X<110)

Z = (X – mean) / SD

Z = (110 – 100) / 10

Z = 10/10

Z = 1

P(Z<1) = P(X<110) = 0.841345

(by using z-table)

Now, find P(X<90)

Z = (X – mean) / SD

Z = (90 – 100) / 10

Z = -10/10

Z = -1

P(Z<-1) = P(X<90) = 0.158655

(by using z-table)

P(90<X<110) = P(X<110) – P(X<90)

P(90<X<110) = 0.841345 – 0.158655

P(90<X<110) = 0.682689

Required probability = 0.682689

What percentage of the population lies between 900 and 1100?

We are given that the population have mean = 100 and SD = 10.

We have to find P(900<X<1100)

P(900<X<1100) = P(X<1100) – P(X<900)

Find P(X<1100)

Z = (X – mean) / SD

Z = (1100 – 100) / 10

Z = 1000/10

Z = 100

P(Z<100) = P(X<1100) = 1.00

(by using z-table)

Now, find P(X<900)

Z = (X – mean) / SD

Z = (900 – 100) / 10

Z = 800/10

Z = 80

P(Z<80) = P(X<900) = 1.00

(by using z-table)

P(900<X<1100) = P(X<1100) – P(X<900)

P(900<X<1100) = 1.00 – 1.00 = 0.00

Required probability = 0.00