Question

A group of 1000 students wrote an entrance exam for the University of Statistics. The mean score was 62 with a standard deviation of 12. Assuming a Normal Distribution, answer the following questions:

1. What is the probability of a student scoring above 75?
2. What is the probability of a student failing? (i.e. below 50) How many students failed?
3. What is the minimum mark you would need to score to be in the top 10%?
4. What is the minimum mark you would need to score to be in the top 1%?
5. How many people scored below 30?
6. What is the probability of scoring between 60 and 80?

Answer 1

A group of 1000 students wrote an entrance exam for the University of Statistics. The mean score was 62 with a standard deviation of 12. Assuming a Normal Distribution, answer the following questions:

What is the probability of a student scoring above 75?

Z value for 75, z =(75-62)/12 = 1.08

P( x >75) = P( z >1.08)

=0.1401

Excel function used: =1-NORM.S.DIST(1.08,TRUE)

What is the probability of a student failing? (i.e. below 50) How many students failed?

Z value for 50, z =(50-62)/12 = -1

P( x < 50) = P( z < -1)

= 0.1587

Excel function used: =NORM.S.DIST(-1,TRUE)

1000*0.1587 = 158.7

students failed = 159 ( rounded)

What is the minimum mark you would need to score to be in the top 10%?

Z value for top 10% = 1.282

X= mean+z*sd = 62+1.282*12

=77.384

What is the minimum mark you would need to score to be in the top 1%?

Z value for top 1% = 2.326

X= mean+z*sd = 62+ 2.326*12

=89.912

How many people scored below 30?

Z value for 30, z =(30-62)/12 = -2.67

P( x < 30) = P( z < -2.67)

= 0.0038

1000*0.0038 = 3.8

students scored below 30 = 4 ( rounded)

What is the probability of scoring between 60 and 80?

Z value for 60, z =(60-62)/12 = -0.17

Z value for 80, z =(80-62)/12 = 1.5

P( 60<x<80) = P( -0.17<z<1.5)

=P( z <1.5) – P( z < -0.17) =0.9332-0.4325

=0.5007