1 lecture 3 outline 1. chebyshev’s theorem 2. the empirical rule 3. measures of relative standing...

1

Lecture 3

Outline

1. Chebyshev’s Theorem2. The Empirical Rule3. Measures of Relative Standing4. Examples

2

Chebyshev’s Theorem

• Applies to any data set.• At least ¾ of the observations in any data set will fall

within 2s (2 standard deviations) of the mean:• ( – 2s, + 2s).

• At least 8/9 of the observations in any data set will fall within 3s of the mean:

• ( – 3s, + 3s).• k >1, at least 1 – (1/k2) of the observations will fall

within ks of the mean. (Works with , too)

Lecture 3

xx x

x x

3

The Empirical Rule

• Applies to distributions that are mound-shaped and symmetric• 68% of the observations will fall within +/- 1s of the mean• 95% of the observations will fall within +/- 2s of the mean• 99.7% (essentially all) of the observations will fall within +/- 3s of the mean• Works with , too.

Lecture 3

4

The Empirical Rule

• Example: IQ scores

o Population mean: = 100o Population std. dev: = 15

o The empirical rule tells us that 99.7% of IQ scores fall in the range 55 to 145.

o Do you see why?

Lecture 3

5

Measures of Relative Standing

A. PercentilesB. Standard scores (Z scores)

Lecture 3

6

Measures of Relative Standing

• Measures of relative standing tell us something about a given score by reporting how it relates to other scores.

• For example, one such measure tells us what proportion of scores in a data set are smaller than a given score.

Lecture 3

7

Measures of Relative Standing

A. Percentiles• With observations in a set arranged in order (smallest to largest), Pth percentile is a number such that P% of the observations fall below it.

• In a set of 200 observations, if a number X is larger than 150 of the observations, then X is at the 75th percentile (150/200 = 75%).

Lecture 3

8

Measures of Relative StandingB. Standard Scores (aka Z scores)

i. For a sample: Z = X – s

ii. For a population: Z = X –

• Z-scores can be positive or negative.• A (raw) score below the mean has a negative Z score.

Lecture 3

x

9

2 Important qualities of the Z score

1. Z-scores are like a ruler – they measure distances. o Z scores give the distance between any score X

and the mean, expressed in standard deviation units.

2. Values expressed in Z scores can be compared, regardless of their original units.

Lecture 3

10

Z scores are like a ruler

• Compare these situations:

= 100 and = 10 = 100 and = 40120 = + 2 120 = + ½

A score of 120 is more impressive on the left, where it is 2 standard deviations above the mean (vs. one-half s on the right).

Lecture 3

11

Z-Scores are unit-free

• Values expressed in Z scores can be compared, regardless of their original units.o For example, suppose you had exam grades for 100

students in Psych 281 and also knew how many hours each student studied for that exam.

o You could compare any student’s Z-score for their grade with the Z-score for their # of hours of studying.

Lecture 3

12

Z-Scores

• Suppose we have the following data for our class:

Mean grade for the exam: 70Standard deviation: 10

Mean # hours studying for exam: 8Standard deviation: 2

Lecture 3

13

Z-Scores

Bill Bob BenGrade 80 80 60Z +1 +1 -1

# hours 6 12 6Z -1 +2 -1

Lecture 3

14

Example – Assignment 2, Q.1

• The distribution of a sample of 100 test scores is symmetrical and mound-shaped, with a mean of 50 and a variance of 144.

a. Approximately how many scores are equal to or greater than 74?

Lecture 3

15

Lecture 3

Example – Assignment 2, Q.1

50 62 74

Since s2 = 144, s = 12, and 74 = + 2s

What percentage of scores falls above the red line?

x

16

Example – Assignment 2, Q. 1

• Since 74 = + 2s, we have p = .975.

• 2.5% of scores are ≥ 74.

• Since there are 100 scores, 2.5% = approx. 3 scores.

Lecture 3

x

17

Example – Assignment 2, Q. 1

b. What score corresponds to the 75th percentile?

• To answer this, we use interpolation. o A score at the 84th percentile is 1 s above the

mean.o = 50 and s = 12, so 62 is one s above the mean –

which is the 84th percentile. (Why?)

Lecture 3

x

18

Example – Assignment 2, Q.1

• Remember that the distribution is mound-shaped and symmetric.

• By the Empirical Rule, 68% of the distribution is between -1 s and +1 s around the mean – so half of that (34%) is between the mean and 1 s above the mean.

Lecture 3

19

Example – Assignment 2, Q. 1

• The 84th percentile is 1 s above the mean by the Empirical Rule.

• A score of 62 is 1 s above the mean because the mean is 50 and s = 12.

• Therefore 84th percentile = 62

Lecture 3

20

Lecture 3

Example – Assignment 2, Q.1

X = 50 X = 62

34%

84th percentile

50% + 34% = 84%

50%

21

Example – Assignment 2, Q. 1

• Now we know which score is at the 50th percentile (the mean score – by definition in a symmetric distribution).

• We also know which score is at the 84th percentile.

• Now we can answer our question: which score is at the 75th percentile?

Lecture 3

22

Example – Assignment 2, Q. 1X %ile

50 50

a 75

62 84

Lecture 3

12

25

34

∆

∆ = 2512 34

23

Example – Assignment 2, Q. 1

∆ = 2512 34

∆ = 25 (12) = 8.82 34

75th percentile:X = 50 + 8.82 = 58.82 ~ 59

Lecture 3

Interpolation

24

Example – Assignment 2, Q. 1

c. If the lowest and highest scores in this sample are 14 and 89, respectively, what is the range of the scores in standard deviation units?

Z1 = 14 – 50 Z2 = 89 – 50

12 12 = -3 .0 = 3.25

Range = -3 .0 to 3.25 = 6.25

Lecture 3

25

Example – Assignment 2, Q. 1

d. Assume the 100 scores have same mean and variance but now have a strongly skewed distribution. At least how many of the scores fall between 32 and 68 in this distribution?

By Chebyshev’s Theorem:

Z1 = 32 – 50 = –1.5 Z2 = 68 – 50 = 1.5

12 12

Lecture 3

26

Example – Assignment 2, Q. 1

k = 1.5

1 – (1/k2) = 1 – (1/1.52) = 1 – 1/2.25

= .555.

55.5% or ~ 56% of scores lie between 32 and 68.

Lecture 3