DESCRIPTIVE STATISTICS

A. Measures of central tendency

I. Simple discrete data

Discrete: Is the data that can be counted

Example: number of children

Number of cars

(we cannot say 1.5 children or 2.4 cars)

Example:

Consider the numbers 6 8 3 2 6 5 1 6 8. Find mean, median, mode

6 8 3 2 6 5 1 6 8 455

9 9

6 (occurs 3 times)

1 2 3 5 6 6 6 8 8

6 is in t

M

he middle

So media

ean

Mode

Medi

n

n

i

a

s 6

Or

Median=

So the 5th

term is the median: number 6

Mean The sum of all

squares divided

by the number of

values

Median The value that

is in the middle

Mode The value that is the

most frequent

Very important!!!!

The numbers must be

in order.

Otherwise we cannot

find the median

What if we had 10 numbers instead of 9?

Example:

1 2 3 5 6 8 8 9 9 9

Now the median is between 6 and 8

So median=

Or

Median=

So the median is between 5th and 6th term: number 7

A hard question:

(a)

7 4 5 8 T 14 47

8

46 T7

8

46 T 56

T 56 46

T 10

(b) Mode=4 (occurs 3 times)

(c) Median: The numbers must be in order

4 4 4 5 7 8 10 14

Median=

(8+1)/2=4.5 so is between 4th

and 5th

term

So median=6

II. Discrete data in a frequency table

This is the same as if the frequency table was like the table below:

Number of

errors

Number of

pages

0 28

1 24

2 20

3 17

4 11

Total 100

a. Discrete

b. Mean =

0x28 1x24 2x20 3x17 4x111.59 (or byGDC)

100

c. Median : 1 ( GDC)

d. Mode : 0 because zero is the most frequent number

here

III. Grouped discrete or continuous data

Continuous data : is the data that can be measured

for example weight , height

Mean: to calculate the mean we need midpoints

Median : This is estimated middle value calculated using the cumulative frequency graph.

Mode : Now we have modal group This is the group class interval that has the largest

frequency

Example

Number of errors Number of

pages

40 ≤t < 50 (45) 20

50 ≤t < 60 (55) 61

60 ≤t < 70 (65) 83

70 ≤t < 80 (75) 90

80 ≤t < 90 (85) 106

90 ≤t < 100 (95) 62

100 ≤t < 110 (105) 49

110 ≤t < 120 (115) 29

Total 500

Mean = 45x20 55x61 ... 115x29

79.6min500

Modal group = 80 t <90

Median= we find the median from the cumulative

frequency graph (we will see an example in the

following pages)

Measures of dispersion

Range: Is the difference between the largest and the smallest value

Quartiles: Quartiles separate the data into four equal parts. Each of these parts contain

25% of the data

1

1( 1)

4

thQ n term Lower Quartile

3

3( 1)

4

thQ n term Upper Quartile

3 1Q Q IQR is the Interquartile range

Box and whisker plots: Is a graphic way to display the median, quartiles and extremes of a

data set on a numbered line to show the distribution of the data.

Outlier: Is any value at least 1.5xIQR from the nearest quartile (lower or upper)

Variance: Is the arithmetic mean of the squared differences between each value and the

mean value and is calculated using GDC only

Standard deviation: Is the square root of variance. Low standard deviation shows that the

data points tend to be very close to the mean and high standard deviation shows that the

data is spread out over a large range of values.

Range

Quartiles

Standard deviation

Variance

Example: Simple discrete data

20 23 23 25 26 29 30 30 31

Q1 MEDIAN Q3

Range : 31- 20 = 11

1

n 1 9 1 10Q 2.5

4 4 4

So Q1 is between 23 and 23 → Q1 = 23

Median : (n+1)/2 =10/2 = 5 so the 5th

number

is the median = 26

Q3 = 3 ( n+1) / 4 = 3 (9 + 1 ) / 4 = 30 / 4 = 7.5

So Q3 is between the 7th

and the 8th

number

Q3 = 30

Q3 - Q1 = I Q R = 30 - 23 = 7

Mean =20 23 23 ... 31

26.39

Variance = 2 23.65148371 13.3 (GDC)

Standard deviation= 3.65 (GDC)

STEM A& LEAF DIAGRAMS

Example: Find the median weight in kg of 34 eight year-old children

Median weight: (n+1) / 2 = (34 + 1)/2 = 17.5

So we must find the number that is

between the 17th

and 18th

term

(2+4) /2 = 3 median 30.3 kg

Example

b. i. Median: (n+1)/2 = (19 + 1)/2 = 10th term 88cm

ii. Q1 = (n+1)/4 = (19 + 1)/4 = 5th term 78 cm

Q3 = 3( n+1)/4 = 3 (19+1)/ 4 = 15th term 103 cm

CUMULATIVE FREQUNCY GRAPH

An example of how we use and what can be asked in a cumulative frequency graph is given

below:

50(a) 80 40

100

Because 50/100 means the middle

So median = 30

(b) IQR = Q3-Q1

11

33

25Q 80 20

100

75Q 80

So

6

Q 20

So Q 450100

Q3-Q1=45 – 20=25

thSo 85 percentile=23

50 or ab

35(c) 80 28

100

10(d) 100% 12.5%

80ove: