1. Probability & the Normal Distribution
Tuesday, September 3, 2013
Probability & the Normal Distribution

2. First, some loose ends from last time

3. Other standardized distributions57 29 43 85
= =14

4. Other standardized distributions57 85 29 43 -2 -1 1 2
Original (X): = =14
Z-Scores: =0 =1

5. Other standardized distributions57 85 29 43 -2 -1 1 2 30 40 50 60 70
Original (X): = =14
Z-Scores: =0 =1
Standardized: = =10

6. In-Class Exercise:
Find the standard deviation for the following population of scores: 1,3,4,4,5,7,9
Find the standard deviation for the following sample of scores: 1,2,2,3,9,10
For a distribution with µ=40 and =12, find the z-score for each of the following scores: a. X=36 b. X=46 c. X=56
A population with a mean of µ=44 and a standard deviation of =6 is standardized to create a new distribution of with µ=50 and =10.
What is the new value for an original score of X=47?
If the new score is 65, what was the original score?

7. In-Class Exercise:
Find the standard deviation for the following population of scores: 1,3,4,4,5,7,9
Find the standard deviation for the following sample of scores: 1,2,2,3,9,10
For a distribution with µ=40 and =12, find the z-score for each of the following scores: a. X=36 b. X=46 c. X=56
A population with a mean of µ=44 and a standard deviation of =6 is standardized to create a new distribution of with µ=50 and =10.
What is the new value for an original score of X=47?
If the new score is 65, what was the original score?

8. Use z scores to relate the two distributions to each other
Original Distribution: µ=44, =6
Standardized distribution: µ=50, =10
What is the new (standardized) value for an original score of X=47?
Z = (X-μ)/ = (47-44)/6 = +0.5
X = Z  + μ = 0.5* = 55
If the new (standardized) score is 65, what was the original score?
Z = (X-μ)/ = (65-50)/10 = +1.5
X = Z  + μ = 1.5* = 53

9. In-Class Exercise:
Find the standard deviation for the following population of scores: 1,3,4,4,5,7,9
Find the standard deviation for the following sample of scores: 1,2,2,3,9,10
For a distribution with µ=40 and =12, find the z-score for each of the following scores: a. X=36 b. X=46 c. X=56
A population with a mean of µ=44 and a standard deviation of =6 is standardized to create a new distribution of with µ=50 and =10.
What is the new value for an original score of X=47?
If the new score is 65, what was the original score?

10. Use z=(x-μ)/ µ=40 and =12, X=36 Z = (36-40)/12 = -4/12 = -0.33 X=46

11. In-Class Exercise:
Find the standard deviation for the following population of scores: 1,3,4,4,5,7,9
Find the standard deviation for the following sample of scores: 1,2,2,3,9,10
For a distribution with µ=40 and =12, find the z-score for each of the following scores: a. X=36 b. X=46 c. X=56
A population with a mean of µ=44 and a standard deviation of =6 is standardized to create a new distribution of with µ=50 and =10.
What is the new value for an original score of X=47?
If the new score is 65, what was the original score?

12. Use formula for s (sample SD): s=
1,2,2,3,9,10
M = 27/6 = 4.5
X-M = -3.5, -2.5, -2.5, -1.5, 4.5, 5.5
(X-M)2 = 12.25, 6.25, 6.25, 2.25, 20.25, 30.25
(X-M)2 = 77.5
s2=77.5/(n-1) = 77.5/5 = 15.5
s = 3.94
Or use Excel stdev.s command

13. Today: Probability & the Normal Distribution
Any questions from last time?

14. Topics for today Review of probability (Chapter 6)
Binomial Distribution
Normal Distribution

15. Basics of Probability Probability Outcome
Expected relative frequency of a particular outcome, in a situation in which several different outcomes are possible
Outcome
Could be the result of a coin toss or experiment, could be obtaining a particular score on a variable of interest 15

16. Flipping a coin example
What are the odds of getting a “heads”?
n = 1 flip
One outcome classified as heads = 1 2
Total of two outcomes =
0.5 16

17. Flipping a coin example
What are the odds of getting two “heads”?
n = 2
Number of heads 2
Four total outcomes
One 2 “heads” outcome 1 =
0.25 1
This situation is known as the binomial
# of outcomes = 2n 17

18. Flipping a coin example
What are the odds of getting “at least one heads”?
n = 2
Number of heads 2
Four total outcomes
Three “at least one heads” outcome 1 =
0.75 1 18

19. Flipping a coin example
Number of heads
n = 3
HHH 3
HHT 2
HTH 2
HTT 1 2
THH
THT 1
TTH 1
TTT 2n
= 23 = 8 total outcomes 19

20. HHH 5 3H = 5
HHH
HHT 5 2H = 11
HHT
HTH 3
HTH
THH 3
THH
HTT 0 1H = 4
THT 2
THT
TTH 2
TTH
TTT 6 0H = 6
TTT

21. Connection between probabilities & graphs
We usually have a population of scores that can be displayed in a graph (such as a histogram)
Each portion of the graph represents a different proportion of the population
The proportion is equivalent to the probability of obtaining an individual in that portion of the graph

22. Example
Population with the following scores: 1,1,2,3,3,4,4,4,5,6

23. Example What is the probability of obtaining a score greater than 4?
p(X>4) = ?

24. Example Find the following probabilities: p(X>2) = ? p(X>5) = ?

25. Check your understanding
We are about to look at the normal distribution and see how probability concepts are related to this specific distribution.
Before we move on, any questions about probability, how to compute it, how it is related to frequency graphs, etc.?

26. The Normal Distribution

27. The Normal Distribution
Normal distribution is a commonly found distribution that is symmetrical and unimodal.
Not all unimodal, symmetrical curves are Normal, so be careful with your descriptions
It is defined by the following equation:
The mean, median, and mode are all equal for this distribution. 1 2 -1 -2

28. The Normal Distribution
This equation provides x and y coordinates on the graph of the frequency distribution. You can plug a given value of x into the formula to find the corresponding y coordinate. Since the function describes a symmetrical curve, note that the same y (height) is given by two values of x (representing two scores an equal distance above and below the mean) 1 2 -1 -2
Y =

29. The Normal Distribution
As the distance between the observed score (x) and the mean increases, the value of the expression (i.e., the y coordinate) decreases. Thus the frequency of observed scores that are very high or very low relative to the mean, is low, and as the difference between the observed score and the mean gets very large, the frequency approaches 0. 1 2 -1 -2
Y =

30. The Normal Distribution
As the distance between the observed score (x) and the mean decreases (i.e., as the observed value approaches the mean), the value of the expression (i.e., the y coordinate) increases.
The maximum value of y (i.e., the mode, or the peak in the curve) is reached when the observed score equals the mean – hence mean equals mode. 1 2 -1 -2
Y =

31. The Normal Distribution
The integral of the function gives the area under the curve (remember this if you took calculus?)
The distribution is asymptotic, meaning that there is no closed solution for the integral.
It is possible to calculate the proportion of the area under the curve represented by a range of x values (e.g., for x values between -1 and 1). 1 2 -1 -2
Y =

32. Check your understanding
Next we will see how probability concepts are related to the normal distribution, by learning about the Unit Normal Table.
Before we move on, any questions about the properties of the normal distribution?

33. The Unit Normal Table (Appendix B)
The normal distribution is often transformed into z-scores. z
Body
Tail :
0.5
1.0
2.8
2.9
0.5000
0.6915
.8413
.9974
.9981
.3085
.1587
.0026
.0019
Unit Normal Table gives the precise proportion of scores (in z-scores) between the mean (Z score of 0) and any other Z score in a Normal distribution
Contains the proportions in the tail to the left of corresponding z-scores of a Normal distribution
This means that the table lists only positive Z scores
Note that for z=0 (i.e., at the mean), the proportion of scores to the left is .5 Hence, mean=median.

34. Using the Unit Normal Tablez
Body
Tail :
0.5
1.0
2.8
2.9
0.5000
0.6915
.8413
.9974
.9981
.3085
.1587
.0026
.0019
50%-34%-14% rule
Similar to the 68%-95%-99% rule
13.59%
2.28%
34.13%
At z = +1: -2 -1 1 2
15.87% (13.59% and 2.28%) of the scores are to the right of the score
100%-15.87% = 84.13% to the left

35. Using the Unit Normal Table
Steps for figuring the percentage above or below a particular raw or Z score: z
Body
Tail :
0.5
1.0
2.8
2.9
0.5000
0.6915
.8413
.9974
.9981
.3085
.1587
.0026
.0019
1. Convert raw score to Z score (if necessary) 2. Draw normal curve, where the Z score falls on it, shade in the area for which you are finding the percentage 3. Make rough estimate of shaded area’s percentage (using 50%-34%-14% rule)

36. Using the Unit Normal Table
Steps for figuring the percentage above or below a particular raw or Z score: z
Body
Tail :
0.5
1.0
2.8
2.9
0.5000
0.6915
.8413
.9974
.9981
.3085
.1587
.0026
.0019
4. Find exact percentage using unit normal table
5. If needed, subtract percentage from 100%.
6. Check the exact percentage is within the range of the estimate from Step 3

37. SAT Example problems The population parameters for the SAT are:
m = 500, s = 100, and it is Normally distributed
Suppose that you got a 630 on the SAT. What percent of the people who take the SAT get your score or lower?
So 90.32% got your score or lower
From the table:
z(1.3) =.9032
That’s 9.68% above this score

38. Check your understanding
Next we will see how to figure out a z score if you know the percentile
Before we move on, any questions about the connection between probabilities and distributions?
Questions about using the unit normal table to find the % of a distribution falling above or below a z score?

39. The Normal Distribution
You can go in the other direction too
Steps for figuring Z scores and raw scores from percentages:
1. Draw normal curve, shade in approximate area for the percentage (using the 50%-34%-14% rule)
2. Make rough estimate of the Z score where the shaded area starts
3. Find the exact Z score using the unit normal table
4. Check that your Z score is similar to the rough estimate from Step 2
5. If you want to find a raw score, change it from the Z score

40. The Normal Distribution
Example: What z score is at the 75th percentile (at or above 75% of the scores)?
1. Draw normal curve, shade in approximate area for the % (use the 50%-34%-14% rule)
2. Make rough estimate of the Z score where the shaded area starts (between .5 and 1)
3. Find the exact Z score using the unit normal table (a little less than .7)
4. Check that your Z score is similar to the rough estimate from Step 2
5. If you want to find a raw score, change it from the Z score using mean and standard deviation info.

41. The Normal Distribution
Finding the proportion of scores falling between two observed scores
Convert each score to a z score
Draw a graph of the normal distribution and shade out the area to be identified.
Identify the area below the highest z score using the unit normal table.
Identify the area below the lowest z score using the unit normal table.
Subtract step 4 from step 3. This is the proportion of scores that falls between the two observed scores. 1 2 -1 -2

42. The Normal Distribution1 2 -1 -2
Example: What proportion of scores falls between the mean and .2 standard deviations above the mean?
Convert each score to a z score (mean = 0, other score = .2)
Draw a graph of the normal distribution and shade out the area to be identified.
Identify the area below the highest z score using the unit normal table:
For z=.2, the proportion to the left = .5793
Identify the area below the lowest z score using the unit normal table.
For z=0, the proportion to the left = .5
Subtract step 4 from step 3:
= .0793
About 8% of the observations fall between the mean and .2 SD.

43. The Normal Distribution1 2 -1 -2
Example 2: What proportion of scores falls between -.2 standard deviations and -.6 standard deviations?
Convert each score to a z score (-.2 and -.6)
Draw a graph of the normal distribution and shade out the area to be identified.
Identify the area below the highest z score using the unit normal table:
For z=-.2, the proportion to the left = = .4207
Identify the area below the lowest z score using the unit normal table.
For z=-.6, the proportion to the left = = .2743
Subtract step 4 from step 3:
= .1464
About 15% of the observations fall between -.2 and -.6 SD.

44. Check your understanding
Next we will see how the shape of the binomial distribution is similar to that of the normal distribution.
Before we move on, any questions about use of the unit normal table?

45. Flipping a coin example
Number of heads
HHH 3
HHT 2
HTH 2
HTT 1 2
THH
THT 1
TTH 1
TTT 2n
= 23 = 8 total outcomes 45

46. Flipping a coin example
Number of heads
Distribution of possible outcomes
(n = 3 flips) 3 2
Number of heads 1 2 3 .1 .2 .3 .4
probability X f p 3 1
.125 2
.375 2
.375
.375 1 2
.125
.125 1 1 46

47. Flipping a coin example
Can make predictions about likelihood of outcomes based on this distribution.
Distribution of possible outcomes
(n = 3 flips) .4
What’s the probability of flipping three heads in a row? .3
probability .2 .1
.125
p = 0.125
.375
.375
.125 1 2 3
Number of heads 47

48. Flipping a coin example
Can make predictions about likelihood of outcomes based on this distribution.
Distribution of possible outcomes
(n = 3 flips) .4
What’s the probability of flipping at least two heads in three tosses?
probability .3 .2 .1
.125
p = = 0.50
.375
.375
.125 1 2 3
Number of heads 48

49. Flipping a coin example
Can make predictions about likelihood of outcomes based on this distribution.
Distribution of possible outcomes
(n = 3 flips) .4
What’s the probability of flipping all heads or all tails in three tosses?
probability .3 .2 .1
.125
p = = 0.25
.375
.375
.125 1 2 3
Number of heads 49

50. Binomial Distribution
Two categories of outcomes (A, B) (e.g., coin toss)
p=p(A) = Probability of A (e.g., Heads)
q=p(B) = Probability of B (e.g., Tails)
p + q = 1.0 (e.g., – could be different values)
n = number of observations (e.g., coin tosses)
X = number of times category A occurs in a sample
If pn > 10 and qn > 10, X follows a nearly normal distribution with μ = pn and σ =

51. HHH 5 3H = 5
HHH
HHT 5 2H = 11
HHT
HTH 3
HTH
THH 3
THH
HTT 0 1H = 4
THT 2
THT
TTH 2
TTH
TTT 6 0H = 6
TTT