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 andpsychological 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
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.
b. Z = (560 - 500) / 100 = 0.60
c. Area between mean and Z is .2257
d. Answer is .2257
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.
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?
a.
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
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.
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.Answer is total area minus the area between mean and 0.5= .4772 - .1915 = .2857
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?
a.
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
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.
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.
Exercise 7. A college decides to admit only the top 10% of SAT students. What would its cutoff be?
a.
b. Z score corresponding to Col 2 = .3997 orCol 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?
a.
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?
a.
b. 2nd Magic Z is 2.58.
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?
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?
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?
Z = (87 – 70) / 10 = 1.70
Answer = .5000 + .4554 = .9554 or Percentile rank = 95.54
Hints/answers
1. Z90 = (90-100)/15 = -.67Z110 = (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) - 111/4/2018