Topic 7: The Normal Distribution
Random variable
A variable whose values are taken on randomly. A variable whose values we don’t know
ahead of time.
Examples:
The number of cars in line at the vehicle emissions test station when you arrive.
The length of time from when you’re seated until the server takes your drink order.
The amount you spend on transportation each year.
Probability Distribution
A table or formula which gives the probabilities of various values of some random
variable.
Some common Distributions
Heights of randomly selected adult males.
Diameters of randomly selected Oak trees.
Time to wait in checkout lane.
Fact: Thousands of physical and psychological measures all have distributions with the
same essential shape . . .
This shape is so ubiquitous that a couple of 100 years ago, mathematicians began trying
to discover a formula that characterized it.
The formula that was discovered is
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 1
2/6/2016
Normal Distribution In-class exercises
Two types of problems you'll encounter involving the Normal Distribution:
1. Given X or X's, find Probability (or percentage).
Example: What is the percentage of IQ scores between 100 and 120?
a. Draw a picture and shade the area corresponding to the probability to be found
b. Convert the X(s) to Z(s).
c. Look up probability or probabilities corresponding to Z or Z's in Table A
Appendix C.
d. Compute the answer from the tabled probabilities.
2. Given Probability or percentage, find X or X's.
Example: What is the IQ score below which 60% of all IQ's fall?
a. Draw a picture. Label only the mean. Draw line(s) roughly corresponding to
X's.
b. Find Z or Z's corresponding to probability or probabilities in Table A.
c. Convert Z(s) to X(s) using the inverse of the Z formula, X = Z*Sigma + Mu.
For the following exercises, assume we're working with normally distributed SAT scores.
Mu = 500
Sigma = 100
Exercise 1. A student is randomly selected from the SAT population. What's the
probability of that student's score being between 500 and 560?
a.
300
400
500
600
700
0
b. Z = (560 - 500) / 100 = 0.60
c. Area between mean and Z is .2257
d. Answer is .2257
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 2
2/6/2016
Exercise 2. A student is randomly selected from the SAT population. What's the
probability of that student's score being between 440 and 500?
a.
300
400
500
600
700
b. Z = (440-500)/100 = -0.60
c. Area in Table C is .2257
d. Answer is also .2257
Exercise 3. A student is randomly selected from the SAT population. What's the
probability of that student's score being between 350 and 600?
.3413
.4332
a.
300
400
500
600
700
b.
Z350= (350-500)/100 = -1.50, that is, minus 1.50
Z600 = (600-500)/100 = 1.00
c.
Table A for Z of 1.50 = .4332
Table A for Z of 1.00 = .3413
d.
Answer is .4332 + .3413 = .7745
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 3
2/6/2016
Exercise 4. A student is randomly selected from the SAT population. What's the
probability of that student's score being between 550 and 700?
a.
300
400
500
600
700
b.
Z700 = (700-500)/100 = 2.00 (The total area.)
Z550 = (550-500)/100 = 0.50 (The amount to be sold off.)
c.
For Z = 2.00, Table A area between mean and Z = .4772
For Z = 0.50, Table A area between mean and Z = .1915
d.
= .2857
Answer is total area minus the area between mean and 0.5= .4772 - .1915
Exercise 5. A student is randomly selected from the SAT population. What's the
probability of that student's score being between 250 and 350?
.4938
.4332
a.
300
400
500
600
700
b.
For X=250, Z = (250-500) / 100 = -2.50
For X=350, Z = (350-500) / 100 = -1.50
c.
For Z = 2.50, area between the mean and Z is .4938
For Z = 1.50, area between the mean and Z is .4332
d. We sell off the area between mean and 1.50, getting rid of .4332.
The remainder is .4938 - .4332 = .0606
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 4
2/6/2016
Exercise 6. A college decides to admit students with SAT scores greater than or equal to
450. Assuming the applicant population contains 1500 students, how many would be
admitted?
a.
300
400
500
600
700
b. For X = 450, Z = (450-500) / 100 = -0.50
c. For Z = 0.50, area between mean and Z is .1915.
I know because the Normal distribution is symmetric that the area above
the mean is .5000.
d. Pre-answer is the sum of the area between 450 and 500, .1915 and the area
above 500, which is .5000. The pre-answer is .6915.
But wait!! The problem doesn’t want a proportion, it wants the number expected from
1500 applicants. So I have to multiply .6915 * 1500 to get the expected number of
applicants, 1037. So answer is 1037.
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 5
2/6/2016
Exercise 7. A college decides to admit only the top 10% of SAT students. What would
its cutoff be?
.1000
.5000
.4000
a.
500
b. Z score corresponding to Col 2 = .3997 or Col 3 = .1003 is 1.28.
Z = (X-Mean)/S ~~ Z*S = X-Mean ~~ Z*S + Mean = X
c. Answer : Z = 1.28*100 + 500 = 628.
Exercise 8. What are the two Z-scores equal distance from the mean such that 95% of the
scores are between them?
95%
47.5%
47.5%
a.
0
Z=-1.96
Z=+1.96
b.
c. We could find the values, but I’m only interested here in the Z-scores.
Exercise 9. What are the two Z-scores equal distance from the mean such that 1% of the
scores are beyond them?
.0050
.0050
.4950
.4950
a.
b. 2nd Magic Z is 2.58.
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 6
2/6/2016
Normal Distribution Drill and practice assignment.
1. Scores on a standard intelligence test are approximately normally distributed with
mean = 100 and standard deviation = 15. What proportion of persons will score between
90 and 110 on that test?
2. Scores on a measure of conscientiousness are approximately normally distributed with
mean = 3.5 and standard deviation = 1. A company decides to hire only persons who
score above 5 on the conscientiousness scale. What proportion of applicants will it hire?
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 7
2/6/2016
3. The ACT scores of students admitted to a college are approximately normally
distributed with mean = 21 and standard deviation = 4. The college is considering special
funding for persons with ACT greater than 30. What proportion of the students attending
the college would be eligible for the special funding?
.0122
21
25
29
Z = (30 – 21) / 4 = 9 /4 = 2.25
4. A student took a statistics test. His instructor said that the test scores were
approximately normally distributed with mean = 70 and standard deviation = 10. The
student scored 87. What was her percentile rank?
.0446
.4554
70
Z = (87 – 70) / 10 = 1.70
Answer = .5000 + .4554 = .9554 or Percentile rank = 95.54
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 8
2/6/2016
Hints/answers
1. Z90 = (90-100)/15 = -.67
Z110 = (110-100)/15 = +.67
Area between mean and Z is .2486 for both. Answer is 2*.2486
2. Z5 = (5-3.5)/1 = 1.5.
Area beyond 1.5 is .0668. So company could hire .0668 = 6.68% of applicants.
3. Z30 = (30-21)/4 = 9/4 = 2.25
Proportion beyond Z of 2.25 is .0122. So college needs funding for .0122 or 1.22% of
students.
4. Z87 = (87-70)/10 = 17/10 = 1.70
Proportion below Z of 1.7 is .5 + proportion between mean and 1.7 = .5 + .4554 = .9554.
So percentile rank of the student is 95.54.
Biderman’s 201 Handouts Topic 7 (Normal Distribution) - 9
2/6/2016