tuesday, september 3, 2013

Tuesday, September 3, 2013

Probability & the Normal Distribution

First, some loose ends from last time

Other standardized distributions

57 8543

29

=57 =14


57 85-1

-2

Original (X): =57 =14

243

29

Z-Scores: =0 =1

10


57 85-1

-2

Original (X): =57 =14

243

29

Z-Scores: =0 =1

10

Standardized: =50 =10

40

30

706050

In-Class Exercise:1. Find the standard deviation for the following population

of scores: 1,3,4,4,5,7,92. Find the standard deviation for the following sample of

scores: 1,2,2,3,9,103. 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

4. 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. a. What is the new value for an original score of X=47?

b. If the new score is 65, what was the original score?

Use z scores to relate the two distributions to each other

Original Distribution: µ=44, =6

Standardized distribution: µ=50, =10

a. 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*10 + 50 = 55

b. 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*6 + 44 = 53

Use z=(x-μ)/

µ=40 and =12,

X=36

Z = (36-40)/12 = -4/12 = -0.33

X=46

Z = (46-40)/12 = 6/12 = +0.5

X=56

Z = (56-40)/12 = 16/12 = +1.33

Use formula for s (sample SD): s=

1,2,2,3,9,10M = 27/6 = 4.5X-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, 30.25S (X-M)2 = 77.5s2=77.5/(n-1) = 77.5/5 = 15.5s = 3.94

Or use Excel stdev.s command

Today: Probability & the Normal Distribution

Any questions from last time?

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

Basics of Probability

• Probability– 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

Flipping a coin example

What are the odds of getting a “heads”?

One outcome classified as heads=

1

2=0.5

Total of two outcomes

n = 1 flip


What are the odds of getting two “heads”?

Number of heads

2

1

1

0

One 2 “heads” outcomeFour total outcomes

=0.25

This situation is known as the binomial# of outcomes = 2n

n = 2


What are the odds of getting “at least one heads”?

Number of heads

2

1

1

0

Four total outcomes

=0.75

Three “at least one heads” outcome

n = 2


HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

Number of heads3

2

1

0

2

2

1

1

2n= 23 = 8 total outcomes

n = 3

HHH 5 3H = 5HHHHHHHHHHHH HHT 5 2H = 11HHTHHTHHTHHT HTH 3HTHHTH THH 3THHTHH HTT 0 1H = 4

THT 2THT TTH 2TTH TTT 6 0H = 6TTTTTTTTTTTTTTT

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

Example

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

Series10

1

2

3

4

2

1

2

3

1 1

1 2 3 4 5 6

Example

• What is the probability of obtaining a score greater than 4?

• p(X>4) = ?

Series10

1

2

3

4

2

1

2

3

1 1

1 2 3 4 5 6

Example

Find the following probabilities:• p(X>2) = ?• p(X>5) = ?• P(X<3) = ?

Series10

1

2

3

4

2

1

2

3

1 1

1 2 3 4 5 6

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.?

The Normal Distribution

• Normal distribution

http://en.wikipedia.org/wiki/Normal_distribution

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 0


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 0

Y =


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 0

Y =


• 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 0

Y =

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 0

Y =


• 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?

The Unit Normal Table (Appendix B)

• 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.

The normal distribution is often transformed into z-scores. z Body Tail

0::

0.5::

1.0::

2.82.9

0.5000::

0.6915::

.8413::

.9974

.9981

0.5000::

.3085::

.1587::

.0026

.0019

Using the Unit Normal Table

z Body Tail

0::

0.5::

1.0::

2.82.9

0.5000::

0.6915::

.8413::

.9974

.9981

0.5000::

.3085::

.1587::

.0026

.0019 15.87% (13.59% and 2.28%) of the scores are to the right of the score100%-15.87% = 84.13% to the left

At z = +1:

13.59%2.28%

34.13%

50%-34%-14% rule

1 2-1-2 0

Similar to the 68%-95%-99% rule


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)

• Steps for figuring the percentage above or below a particular raw or Z score:

z Body Tail

0::

0.5::

1.0::

2.82.9

0.5000::

0.6915::

.8413::

.9974

.9981

0.5000::

.3085::

.1587::

.0026

.0019


4. Find exact percentage using unit normal table5. If needed, subtract percentage from 100%. 6. Check the exact percentage is within the range of the estimate from Step 3

• Steps for figuring the percentage above or below a particular raw or Z score:

z Body Tail

0::

0.5::

1.0::

2.82.9

0.5000::

0.6915::

.8413::

.9974

.9981

0.5000::

.3085::

.1587::

.0026

.0019

Suppose that you got a 630 on the SAT. What percent of the people who take the SAT get your score or lower?

SAT Example problems

• The population parameters for the SAT are: m = 500, s = 100, and it is Normally

distributed

From the table:

z(1.3) =.9032 That’s

9.68% above this score

So 90.32% got your score or lower


• 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?


• 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 starts3. Find the exact Z score using the unit normal table4. Check that your Z score is similar to the rough estimate from Step 25. If you want to find a raw score, change it from the Z score

The Normal DistributionExample: 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 25. If you want to find a raw score, change it from the Z score using mean and standard deviation info.


Finding the proportion of scores falling between two observed scores

1. Convert each score to a z score2. Draw a graph of the normal distribution and shade out the area to be

identified.3. Identify the area below the highest z score using the unit normal table.4. Identify the area below the lowest z score using the unit normal table.5. Subtract step 4 from step 3. This is the proportion of scores that falls

between the two observed scores.

1 2-1-2 0


Example: What proportion of scores falls between the mean and .2 standard deviations above the mean?

1. Convert each score to a z score (mean = 0, other score = .2)

2. Draw a graph of the normal distribution and shade out the area to be identified.

3. Identify the area below the highest z score using the unit normal table:For z=.2, the proportion to the left = .5793

4. Identify the area below the lowest z score using the unit normal table.For z=0, the proportion to the left = .5

5. Subtract step 4 from step 3: .5793 - .5 = .0793

About 8% of the observations fall between the mean and .2 SD.

1 2-1-2 0


Example 2: What proportion of scores falls between -.2 standard deviations and -.6 standard deviations?

1. Convert each score to a z score (-.2 and -.6)2. Draw a graph of the normal distribution and shade out

the area to be identified.3. Identify the area below the highest z score using the unit

normal table:For z=-.2, the proportion to the left = 1 - .5793 = .4207

4. Identify the area below the lowest z score using the unit normal table.For z=-.6, the proportion to the left = 1 - .7257 = .2743

5. Subtract step 4 from step 3: .4207 - .2743 = .1464

About 15% of the observations fall between -.2 and -.6 SD.

1 2-1-2 0


• 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?


HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

Number of heads3

2

1

0

2

2

1

1

2n= 23 = 8 total outcomes

Flipping a coin exampleNumber of heads

3

2

1

0

2

2

1

1

X f p

3 1 .125

2 3 .375

1 3 .375

0 1 .125Number of heads0 1 2 3

.1

.2

.3

.4

probability

.125 .125.375.375

Distribution of possible outcomes(n = 3 flips)


Number of heads0 1 2 3

.1

.2

.3

.4

probability

What’s the probability of flipping three heads in a row?

.125 .125.375.375 p = 0.125


Can make predictions about likelihood of outcomes based on this distribution.



.1

.2

.3

.4

probability

What’s the probability of flipping at least two heads in three tosses?

.125 .125.375.375 p = 0.375 + 0.125 = 0.50





.1

.2

.3

.4

probability

What’s the probability of flipping all heads or all tails in three tosses?

.125 .125.375.375 p = 0.125 + 0.125 = 0.25



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., .5 + .5 – 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 =

HHH 5 3H = 5HHHHHHHHHHHH HHT 5 2H = 11HHTHHTHHTHHT HTH 3HTHHTH THH 3THHTHH HTT 0 1H = 4

THT 2THT TTH 2TTH TTT 6 0H = 6TTTTTTTTTTTTTTT

Series10

2

4

6

8

10

5

11

4

6

3 Heads 2 Heads 1 Heads

tuesday, september 3, 2013

Documents

original x

original score of x

new score

z scores

new standardized score

following scores

standard deviation

following population