The Normal Distribution

PSYC 6130, PROF. J. ELDER 2

is the mean

is the standard deviation

• The height of a normal density curve at any point x is given by

1 22( )1

( )2

x

f x e

Normal Distribution

Recall:

3.1416

2.7183e

PSYC 6130, PROF. J. ELDER 3

Normal Distribution

• A Normal Distribution has a symmetric, unimodal and bell-shaped density curve.

• The mean and standard deviation completely specify the curve.

• The mean, median, and mode are the same.

PSYC 6130, PROF. J. ELDER 4

Probabilities and the Normal Distribution

Shaded area = 0.683 Shaded area = 0.954 Shaded area = 0.997

68% chance of fallingbetween and

95% chance of fallingbetween and

99.7% chance of fallingbetween and

PSYC 6130, PROF. J. ELDER 5

Z-Scores

• The z-score is a normalized representation of a random variable with zero mean and unit variance.

• z-scores are dimensionless.

• The z-score tells you how many standard deviations a score lies from the mean.

X

z

68.2%

11.5%

PSYC 6130, PROF. J. ELDER 6

The Standard Normal Table: (Appendix z)• A table of areas (probabilities) under the standard normal density curve. The

table entry for each value z is the area under the curve between the mean and z.

-4 -3 -2 -1 0 1 2 3 4z

PSYC 6130, PROF. J. ELDER 7

PSYC 6130, PROF. J. ELDER 8

Grade N Valid 115

Missing 5Percentiles 10 55.0

20 59.430 62.040 65.350 66.760 69.470 73.180 77.990 83.9

100 97.3

68.2%

11.5%

Estimating percentiles using z-scores

• From a frequency table we can directly compute percentiles and percentile ranks.

• If we model the data as normal, we can also calculate percentiles and percentile ranks using z-scores.

X

z

PSYC 6130, PROF. J. ELDER 9

Z-Scores

Example 1. What is the 50th percentile? Grade N Valid 115

Missing 5Percentiles 10 55.0

20 59.430 62.040 65.350 66.760 69.470 73.180 77.990 83.9

100 97.3

50th percentile 0.5 0.p z

68.2%

11.5%

68.2%X

z X

-4 -3 -2 -1 0 1 2 3 4z

PSYC 6130, PROF. J. ELDER 10

Z-Scores

Grade N Valid 115

Missing 5Percentiles 10 55.0

20 59.430 62.040 65.350 66.760 69.470 73.180 77.990 83.9

100 97.3

Example 2. What is the 75th percentile?

68.2%

11.5%

-4 -3 -2 -1 0 1 2 3 4z

75th percentile 0.75 0.67p z

68.2 11.5 0.67 75.9%X

z X z

PSYC 6130, PROF. J. ELDER 11

Z-Scores

Grade N Valid 115

Missing 5Percentiles 10 55.0

20 59.430 62.040 65.350 66.760 69.470 73.180 77.990 83.9

100 97.3

68.2%

11.5%

-4 -3 -2 -1 0 1 2 3 4z

Example 3. What is the percentile rank of a grade of 85%?

85 68.2

1.46 0.9279 92.79%11.5

Xz p

PSYC 6130, PROF. J. ELDER 12

Z-Scores

Grade N Valid 115

Missing 5Percentiles 10 55.0

20 59.430 62.040 65.350 66.760 69.470 73.180 77.990 83.9

100 97.3

68.2%

11.5%

-4 -3 -2 -1 0 1 2 3 4zExample 4. What proportion of students

received a grade of B+ (75-80%)?

12

75 68.20.59 0.2224

11.5

Xz p

11

80 68.21.03 0.3485

11.5

Xz p

Thus proportion of B+ students = 0.3485-0.2224=0.1261 (12.61%)

PSYC 6130, PROF. J. ELDER 13

Z-Scores

Grade N Valid 115

Missing 5Percentiles 10 55.0

20 59.430 62.040 65.350 66.760 69.470 73.180 77.990 83.9

100 97.3

68.2%

11.5%

-4 -3 -2 -1 0 1 2 3 4zExample 4. What proportion of students

received a grade of C+ (65-70%)?

12

65 68.20.28 0.1103

11.5

Xz p

11

70 68.20.16 0.0636

11.5

Xz p

Thus proportion of C+ students

0.0636 0.1103 0.0636 0.1103 0.1739 (17.39%)

PSYC 6130, PROF. J. ELDER 14

Sampling distribution of the mean

Consider a population with mean and standard deviation .

Assume the population follows a normal distribution.

Suppose we take a random sample of size from the population.n

We compute a sample mean .X

What sort of values do we expect to have, relative to ?X

To answer this question, we must consider taking many different random samples,

each of size .n

1 2We then compute many sample means , ,... X X

We can consider the distribution of these means, just as we considered the distribution

of the original scores.

This distribution of sample means is calle sampling distribution of the d the mean.

PSYC 6130, PROF. J. ELDER 15

Sampling Distribution of the Mean

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Properties of the Sampling Distribution of the Mean

The mean of the sampling distribution of the mean is equal to the population mean:

( )E X

The means of samples do not vary as much as the individuals in the population.

The standard deviation of the sampling distribution of

standard

the mea

error o

n

is call f the meed the an .X

The standard error of the mean decreases as the sample size increases:

X

n

n

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Example

• Chest measurements of 5738 Scottish soldiers by Belgian scholar Lambert Quetelet (1796-1874)

– First application of the Normal distribution to human data

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The sample mean has a sampling distribution

Sampling batches of Scottish soldiers and taking chest measurements. Pop mean = 39.8 in, Pop sd = 2.05 in

1

2

3

4

5

6

7

8

9

10

12

11

34 36 38 40 42 44 46

(a) 12 samples of size n = 6Samplenumber

Chest measurement (in.)

From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 1999.

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12 samples of size 24

34 36 38 40 42 44 46

(b) 12 samples of size n = 24Samplenumber

Chest measurement (in.)

1

2

3

4

5

6

7

8

9

10

12

11

From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 2000.

PSYC 6130, PROF. J. ELDER 20

Histograms from 100,000 samples

(c) n = 100

(b) n = 24

393837 40 41 42

393837 40 41 42

393837 40 41 42

0.0

0.5

1.0

1.5

0.0

0.5

1.0

0.0

0.5

Sample mean of chest measurements (in.)

(a) n = 6

Figure 7.2.2 Standardised histograms of the sample means from 100,000 samples of soldiers (n soldiers per sample).

From Chance Encounters by C.J. Wild and G.A.F. Seber, © John Wiley & Sons, 2000.

PSYC 6130, PROF. J. ELDER 21

Example: Height of US Males, Sampling Distribution of the Mean

60 65 70 75 800

0.5

1

1.5

Height of US Males (in)

Pro

bab

ility

pn=1n=9n=100

PSYC 6130, PROF. J. ELDER 22

The Central Limit Theorem

2

2

For any population with mean and variance ,

the sampling distribution of the mean will approach a normal distribution

with mean and variance / as .n n

Note: the original population does not have to be normal for this to hold!!

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Example: a uniform distribution

n=1

n=2

n=10

n=100

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Example: a chi-squared distribution

n=1

n=2

n=10

n=100

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Z-Scores for Groups

In scientific research, we will often compute the mean from

a series of individual measurements.

X

If we know (or hypothesize) that the underlying population distribution

is normal with known parameters and , it is useful to compute

the z-score corresponding to the group mean:

-

X

Xz

PSYC 6130, PROF. J. ELDER 26

Example: IQ Tests

IQs measured by the Wechsler test are approximately normally distributed

with 100, 15.

What is the probability that our class of 13 students would have a mean

IQ greater than 120, assuming the class is a random sample of the population?

n

What assumptions have we made? Are these justified?

PSYC 6130, PROF. J. ELDER 27

Underlying Assumptions

• Population is normally distributed

• Random sampling

– Every sample of size n has the same probability of being selected.

• All individuals have the same probability of being selected.

• Selection of each individual is independent of the selection of all other individuals.

• Technically, sampling should be with replacement, but in Psychology, sampling is normally without replacement.

End of Lecture 3

Sept 24, 2008